ils/essentialsofdescOOhigbrich 


THE  ESSENTIALS 

OF 

DESCRIPTIVE  GEOMETRY 


r 


BY 

F.    G.    HIGBEE,    M.  E. 

Professor  and  Head  of  Department  of  Descriptive  Geometry  and  Drawing^ 
The  State  University  of  Iowa 


FIR  S  T 

EDITION 

FIRST 

THOUSAND 

»     »  •     »  • 
»     »  »»  »  » 

V     •:':•:••• 

^ 


NEW  YORK 

JOHN   WILEY   &    SONS,    Inc. 

London:    CHAPMAN   &  HALL,   Limited 

IQIS 


Copyright,  1915, 

BY 

F.  G.  HIGBEE 


>7 


^ 


Stanbopc  iprcss 

GILSON  COMPANT 
BOSTON,  U.S.A. 


PREFACE 

It  has  been  the  endeavor  of  the  author  in  writing  this  text  to 
include  only  those  portions  of  descriptive  geometry  which  possess 
industrial  utility  and  which  develop  the  qualities  of  mind  so 
essential  in  a  draftsman. 

First  and  foremost  descriptive  geometry  should  aim  to  teach 
projection.  A  draftsman  must  be  able  both  to  read  and  to  write 
drawings  with  faciUty,  and  projection  is  the  very  grammar  of 
the  language  of  the  designer.  Secondly,  descriptive  geometry 
should  aim  to  develop  the  ability  to  solve  problems  concerning 
,the  relations  of  points,  lines,  and  planes.  These  are,  of  course, 
but  elementary  parts  of  all  engineering  structures  as  shown  on 
drawings  and  it  is  important  that  a  draftsman  be  prepared  to 
solve  problems  relating  to  them  directly  on  the  drawing  board. 
Thirdly,  and  perhaps  most  important  of  all,  descriptive  geometry 
should  aim  to  promote  the  ability  to  analyse  a  problem  into  its 
component  parts,  to  reason  from  a  given  set  of  conditions  to  a 
required  set  of  conclusions,  and  to  build  up  from  the  drawing 
a  mental  picture  of  the  object  which  is  there  represented,  for 
without  the  ability  to  analyse,  to  reason,  and  to  visualize  a 
draftsman  is  lacking  in  the  essential  qualifications  of  his  calling. 

For  these  reasons  the  subject  has  been  discussed  from  the 
point  of  view  of  a  draftsman  and  the  essential  relations  of  points, 
lines,  and  planes  have  been  treated  in  the  third  quadrant.  The 
order  of  the  material  has  been  carefully  considered  and  while 
there  is  some  departure  from  traditional  arrangement  it  is  be- 
lieved that  the  selection  of  material  and  its  arrangement  will  be 
found  both  logical  and  conducive  to  a  natural  development  of 
the  subject.  Many  problems  of  a  carefully  graded  character 
and  of  a  practical  '* flavor"  have  been  inserted  at  frequent  inter- 
vals in  the  belief  that  such  work  is  invaluable  in  fixing  principles 
and  promoting  genuine  interest. 

iii 

330427 


iv  PREFACE 

In  the  discussion  on  surfaces  considerable  variation  may  be 
found  from  other  works  on  the  subject  both  in  content  and  treat- 
ment. It  is  believed,  however,  that  the  material  included  is 
broad  enough  in  character  for  practical  purposes;  and  it  is  hoped 
that  the  method  of  treatment,  the  character  of  the  problems, 
and  the  discussion  on  model  making  will  stimulate  interest  in 
this  important  and  useful  part  of  the  subject. 

In  the  preparation  of  this  text  the  author  has  consulted  and 
acknowledges  his  indebtedness  to  the  following  standard  texts: 
Geometric  Descriptive  by  G.  Monge;  Theoretical  and  Practical 
Graphics  by  F.  N.  Willson;  Elements  of  Descriptive  Geometry 
by  Albert  E.  Church;  Elements  of  Descriptive  Geometry  by 
C.  W.  MacCord;  Engineering  Drawing  by  Thos.  E.  French. 

F.  G.  HIGBEE. 

Iowa  City,  Ia.,  Feb.  i,  1915. 


CONTENTS 

CHAPTER  I  Page 

Orthographic  Projection i 

CHAPTER  II 
Profile  Plane ii 

CHAPTER  III 
Assumption  of  Points  and  Lines 21 

CHAPTER  IV 
Ilanes 27 

CHAPTER  V 
Location  of  Points.  Lines,  and  Planes ^^ 

CHAPTER  VI 

Revolution  of  Points 38 

CHAPTER  VII 
Problems  on  the  Line 45 

CHAPTER  VIII 
Problems  on  the  Plane ;. 52 

CHAPTER  IX 
Problems  on  Angles 59 

CHAPTER  X 
Problems  on  Points,  Lines,  and  Planes 63 

CHAPTER  XI 
Surfaces 84 

CHAPTER  XII 
Plane  Surfaces 91 

V 


vi  CONTENTS 

CHAPTER  XIII  Page 

Cylindrical  Surfaces loi 

CHAPTER  XIV 
Conical  Surfaces 115 

CHAPTER  XV 
Intersection  of  Surfaces 131 

CHAPTER  XVI 
Surfaces  of  Revolution 145 

CHAPTER  XVII 
Warped  Surfaces 155 

CHAPTER  XVIII 
Model  Making 179 

CHAPTER  XIX 
Appendix 197 


ESSENTIALS  OF  DESCRIPTIVE 
GEOMETRY 


CHAPTER   I 
ORTHOGRAPHIC   PROJECTION 

1.  Descriptive  Geometry  is  the  science  of  graphic  representa- 
tion. It^is  the  means  by  which  objects  are  shown  on  drawings 
and  by  which  problems  relating  to  these  objects  are  solved. 

2.  When  an  object  such  as  a  machine,  a  bridge,  a  building,  or 
any  elemental  part  of  such  an  engineering  structure,  is  designed 
a  drawing  or  set  of  drawings  of  it  is  made.  Such  a  drawing  is 
not  only  invaluable  to  the  designer  in  recording  his  ideas  step  by 
step  and  in  assisting  him  in  his  design  but  it  is  also  indispensable 
to  the  artisan  who  constructs  the  object.  A  drawing,  as  used  in 
the  engineering  sense,  is  a  complete  set  of  instructions  from  the 
designer  to  the  workman  by  means  of  which  the  workman  may 
reproduce,  in  exact  shape  and  size  and  in  material  and  finish, 
the  identical  object  which  the  designer  represented  on  the 
drawing. 

The  principles  by  which  the  shape  and  size  of  objects,  and  by 
which  problems  relating  to  such  representations  are  graphically 
solved,  are  found  in  descriptive  geometry.  The  art  of  putting 
on  the  drawing  such  additional  information  as  dimensions,  shop 
notes,  finish,  and  other  data  regarding  the  construction  of  the 
object  represented,  is  not  a  part  of  descriptive  geometry  but  is 
included  as  a  part  of  the  art  which  is  generally  known  as  drafting. 

3.  There  are  two  general  systems  by  which  graphic  represen- 
tations are  made.  One  system  has  for  its  purpose  the  represen- 
tation of  objects  as  they  appear,  and  is  called  scenographic  pro- 
jection.   When  an  object  is  looked  at  from  some  particular 


2  ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 

point  of  view  one  can  get  an  idea  of  its  shape  and  size  because  at 
least  three  faces  of  the  object  are  seen,  and  the  effect  of  light  and 
shade  serves  to  bring  out  the  configuration,  and  because  the  eye 
naturally  compares  and  judges  distances.  If  a  representation 
be  made  showing  how  the  object  appears  to  the  observer,  such  a 
drawing  will  be  a  scenographic  projection,  or,  as  it  is  more  com- 
monly called,  a  perspective  drawing. 

There  are  also  a  number  of  other  forms  of  projection  which 
aim  at  the  same  result,  —  that  is,  to  represent  the  object  as  it 
appears,  —  which  may  be  called  pseudo-perspective  drawing. 
Isometric,  cabinet,  and  oblique  projection  are  examples,  but  as 
these  are  used  as  forms  of  perspective  drawing  and  require 
extensive  treatment  they  will  not  be  considered  in  this  text. 

4.  The  other  system  of  projection  has  for  its  purpose  not  the 
representation  of  objects  as  they  appear,  but  the  representation 
of  objects  as  they  actually  are.  This  form  of  representation  is 
called  orthographic  projection;  and  it  is  this  form  of  projection 
which  is  used  in  all  engineering  drawing  where  the  purpose  of  the 
drawing  is  to  convey,  from  the  draftsman  to  the  workman,  in- 
formation and  directions  for  building. 

It  is  with  orthographic  projection  and  the  use  of  orthographic 
projection  as  a  means  of  solving  drawing-board  problems  graphi- 
cally that  this  text  has  to  do. 

5.  In  perspective  drawing  the  eye  of  the  draftsman  is  assumed 
to  be  at  a  definite  distance  from  the  object.  Therefore  the  rays 
of  sight  which  travel  from  the  eye  to  the  corners  and  other  parts 
of  the  object  converge  to  a  point.  If  now  a  plane  be  set  up  be- 
tween the  eye  and  the  object,  as  in  Fig.  i,  and  the  points  where 
the  lines  of  sight  pierce  this  plane  be  found  and  joined,  the  result 
will  be  a  picture  of  the  object  observed.  It  is  obvious  that  the 
picture  will  be  smaller  than  the  object  itself,  because  the  plane, 
which  is  called  the  picture  plane,  is  between  the  eye  and  the 
object. 

It  will  be  observed,  then,  that  in  perspective  drawing  the  size 
of  the  picture  will  vary  inversely  as  the  distance  of  the  picture 
plane  from  the  object;  when  it  is  between  the  object  and  the 
eye  the  picture  will  be  smaller  than  the  object,  and  when  it  is  on 


ORTHOGRAPHIC  PROJECTION 


the  far  side  of  the  object  the  picture  will  be  larger.  This  is 
due,  of  course,  to  the  fact  that  the  lines  of  sight  converge  to 
the  eye. 

It  will  be  observed  that  the  relation  of  the  lines  in  the  object 
to  the  corresponding  lines  in  the  picture  is  not  the  same  for  all 


iri^^— - 


Fig. 


lines.     In  other  words,  some  lines  are  foreshortened  more  than 
others  owing  to  the  position  of  the  eye. 

It  will  also  be  observed  that  in  perspective  more  than  one  face 
of  the  object  is  seen  from  one  point  of  view. 


4  ESSENTIALS   OF  DESCRIPTIVE   GEOMETRY 

6.  In  engineering  drawing  —  or  in  orthographic  projection  — 
the  eye  of  the  draftsman  is  assumed  to  be  at  infinity.  Therefore, 
the  lines  of  sight  do  not  converge  but  are  parallel.     If  now,  as 


>    \ 

\\ 

\ 

\     \ 

To  the  eye  "^ 

Object 

\ 

\ 

' 

\ 

\ 

\ 

X 

Fig.  2. 


in  Fig.  2,  a  plane  be  set  up  between  the  eye  and  the  object  and 
the  points  where  the  lines  pierce  this  plane  be  found  and  joined 
the  result  will  be  a  picture  of  the  object  not  as  it  appears  but  as 
it  actually  is,  and  each  line  of  the  picture  will  be  the  same  length 
as  the  corresponding  Hne  in  the  object. 


ORTHOGRAPHIC   PROJECTION  5 

It  will  be  observed,  therefore,  that  the  size  of  the  picture  does 
not  vary  with  the  position  of  the  picture  plane.  No  matter  where 
the  plane  is  placed  the  size  of  the  resulting  picture  will  be  the 
same.  Of  course,  in  actual  drawing,  objects  are  drawn  to  scale. 
That  is,  some  part  of  an  inch  is  taken  on  the  drawing  to  represent 
an  inch  on  the  object  in  order  that  large  objects  may  be  drawn 
on  convenient  sized  drawing  sheets,  but  the  reduction  is  the  same 
for  all  lines.  If,  for  example,  one  line  in  the  picture  is  drawn  to 
a  scale  of  \"  =  i"  all  lines  will  be  drawn  to  that  same  scale. 

It  will  be  observed,  then,  that  the  relation  of  lines  in  the  object 
to  the  corresponding  lines  in  the  picture  is  the  same  for  all  lines 
and  is  the  same  no  matter  what  distance  the  picture  plane  is  from 
the  object.  If  the  picture  of  a  line  be  measured,  therefore,  it  will 
be  found  to  equal  in  length  the  line  itself. 

It  will  also  be  observed  that  in  orthographic  projection  only 
(Hie  face  of  an  object  is  visible  from  one  point  of  view,  and, 
therefore,  that  to  show  more  than  one  face  more  than  one  pic- 
ture must  be  drawn  and  more  than  one  point  of  view  assumed. 

7.  In  orthographic  projection  the  picture  planes  are  called 
planes  of  projection  and  since  more  than  one  are  required  to  show 
three  dimensions  of  an  object  one  has  been  assumed  vertical  and 
one  has  been  assumed  horizontal,  because  upon  two  planes  an 
object  may  in  general  be  projected  so  as  to  give  a  complete  idea 
of  its  shape  and  size.  The  horizontal  plane  is  called  the  horizon- 
tal plane  of  projection  and  the  view  of  the  object  obtained  by 
projecting  it  upon  this  plane  is  called  the  plan,  or  the  plan  view, 
or  the  H  projection.  The  vertical  plane  is  called  the  vertical  plane 
of  projection  and  the  view  obtained  by  projecting  the  object  upon 
this  plane  is  called  the  elevation,  or  the  front  view,  or  the  V  pro- 
jection. 

8.  Fig.  3  shows  a  picture  of  the  two  planes  of  projection 
with  an  object  ready  to  project  upon  them.  The  line  where 
the  H  and  V  planes  intersect  is  called  the  ground  line,  —  here- 
after denoted  as  G.  L.,  —  and  in  drawings  the  ground  line  is  the 
line  of  reference  which  indicates  the  location  of  the  object  with 
reference  to  the  planes  of  projection. 

To  project  the  object  in  Fig.  4  upon  the  H  plane,  or  to  get  a 


ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 


Horizontal  Plane 


Ground  Line 


plan  view  of  it,  the  eye  of  the  draftsman  is  supposed  to  be  so  far 
above  the  H  plane  that  the  lines  of  sight  are  perpendicular  to  it. 

Therefore,  draw  perpen- 
diculars from  each  corner 
of  the  object  to  the  H 
plane  and  join  the  points 
where  the  perpendiculars 
pierce  H.  To  project 
the  object  upon  the  V 
plane,  or  to  get  an  eleva- 
tion of  it,  the  eye  of  the 
draftsman  is  assumed  to 
be  in  front  of  the  V 
plane  and  so  far  in  front 
that  the  lines  of  sight 
are  perpendicular  to  it. 
Therefore,  draw  perpen- 
diculars from  the  object 


Vertical  Plane 


Fig.  3. 


to  the  V  plane  and  join  the  points  where  the  perpendiculars 
pierce    V.     //    will    be 


H.  Plane 


Plan 


>-'- 


-r 


notedj  therefore,  that  the 
projection  of  a  point 
upon  a  plane  is  the  foot 
of  a  perpendicular  from 
the  point  to  the  plane. 
It  should  be  kept  in 
mind,  however,  that  Fig. 
4  is  merely  a  picture,  not 
an  actual  representation, 
showing  how  projections 
are  made. 

9.  Now  it  will  be  ob- 
served that  the  two 
planes  of  projection 
stand  at  right  angles  to 
each  other,  and  that  if  the  projections  or  views  which  are 
shown  on  these  planes  are  to  be  represented  as  they  appear  and 


vhilii 


Elevation 

/ 
/ 


/ 


V.  Plane 


Fig. 


ORTHOGRAPHIC  PROJECTION  7 

on  one  sheet  of  paper  these  two  planes  must  be  considered 
as  coinciding  with  the  paper  on  which  the  drawing  is  made. 
This  is  accomplished  as  shown  in  Fig.  5  by  revolving  H 
into  coincidence  with  V.  The  H  plane  and  the  V  plane  are  now 
coinciding  with  the  paper,  but  it  will  be  noted  that  the  relation 
of  the  plan  and  the  elevation  with  respect  to  each  other  and  the 


H.  Plane 

\ 
\ 

\ 

\ 

\ 

\ 

\ 
\    Plan 

\ 
\ 
\ 

\ 

/                  j 

^/ 

1 

/ 

i 

/ 

1 

/ 

/ 

i 

L 

■\          r 

Elevation 

Y.  Plane 

Plan 


Elevation 


Fig.  s. 


Fig.  6. 


ground  line  has  in  no  way  been  changed  by  revolving  the  H  plane; 
each  remains  the  same  distance  from  the  ground  line  as  before, 
and  it  will  be  noted  carefully  that  the  plan  and  eleroaiion  of  the 
same  point  lie  in  the  same  perpendicular  to  the  ground  line. 

In  a  drawing  the  planes  of  projection  are  not  Kmited  in  extent, 
so  no  boundary  lines  need  be  shown  for  them.  A  ground  line, 
as  in  Fig.  6,  is  drawn  which  indicates  the  position  of  the  planes 


8 


ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 


with  respect  to  the  object,  —  it  being  understood  that  the  H 
plane  is  above  the  V  plane,  —  and  the  plan  and  elevation  of  the 
object  are  drawn  in  the  required  position  by  locating  them  at  the 
proper  distances  from  the  G.  L.  Fig.  6  shows  a  plan  and  eleva- 
tion of  an  object  as  usually  drawn  in  descriptive  geometry. 

ID.  Now  since  the  two  planes  of  projection  are  indefinite  in 
extent,  and  stand  at  right  angles  to  each  other,  it  will  be  clear 

that  such  a  pair  of 
planes  will  divide 
space  into  four  quar- 
ters as  shown  by  the 
pictorial  drawing  in 
Fig.  7.  These  four 
quarters  are  called 
quadrants,  or  angles, 
and  for  convenience 
are  designated  as 
First,  Second,  Third, 
and  Fourth  Quad- 
rants .  D  ra wings  may 
be  made  with  the  ob- 
ject in  any  of  these 
quadrants,  but  all 
commercial  drawing 
is  done  either  in  the 
third  or  first  quad- 
rant. The  reason  for 
this  will  be  evident 
from  a  study  of  Fig. 
7A  which  shows  how  the  quadrants  are  located  with  respect  to 
the  G.  L.  when  the  planes  of  projection  coincide  with  the  paper. 
From  this  figure  it  will  be  seen  that  in  third  quadrant  drawing 
the  plan  is  above  the  elevation;  in  first  quadrant  drawing  the 
plan  is  below  the  elevation;  and  in  second  and  fourth  quadrant 
drawing  the  plan  and  elevation  are  on  the  same  side  of  the  ground 
line,  and  therefore  often  interfere  with  each  other. 
Numbering  the  quadrants  is  done  to  afford  a  convenient  way 


V 

\ 

\ 
\ 
\ 

V 

\ 
\ 
\ 

A\ 

/^ 

1 
1 
I 
\ 

\ 
\ 

\ 

\ 

\ 

1 
\ 

\ 

\ 

\ 
\ 
\ 
\ 

Fig.  7. 


ORTHOGRAPHIC   PROJECTION  9 

of  indicating  the  location  of  an  object  with  respect  to  the  planes 
of  projection.  When  an  object  is  said  to  be  in  the  first  quadrant 
it  means  simply  that  it  is  above  the  H  plane  and  in  front  of  the 
V  plane;  therefore  its  plan  view  will  be  below  the  elevation. 
When  an  object  is  said  to  be  in  the  third  quadrant  it  means  that 


First  Quadrant.          Second  Quadrant.        Third  Quadrant.  Fourth  Quadrant. 
Plan  below  G.  L.  and   H  and  V  coinciding  Plan  above  G.  L.  and  H  and  V  coinciding 
elevation  above  G.  L.     above  G.  L.  There-  elevation  below  G.  L.  below  G.  L.  There- 
fore plan  and  eleva-  fore  both  views  be- 
tion ho\h.above G.  L.  low  G.  L. 
Fig.  7A. 

it  is  below  H  and  behind  V;  therefore  its  plan  view  will  be 
above  its  elevation.  Also  when  a  drawing  is  examined  these  rela- 
tions serve  to  assist  in  visualizing  the  object,  and  once  the 
''language"  of  projection  and  drawing  is  learned  these  details 
are  not  thought  of  as  such,  and  one  reads  the  drawing  much  as 
one  reads  a  sentence  without  analysing  it  into  subject,  object, 
verb,  etc. 


PROBLEMS  IN  PROJECTION 

1.  Draw  the  plan  and  elevation  of  a  regular  hexagonal  prism  whose  base 
is  2"  in  diameter  and  whose  altitude  is  3''. 

2.  Draw  the  plan  and  elevation  of  a  regular  pentagonal  pyramid  whose 
altitude  is  3"  and  the  sides  of  whose  base  are  2". 

3.  Draw  the  plan  and  elevation  of  a  regular  trimcated  hexagonal  pyra- 
mid whose  lower  base  is  3"  in  diameter,  whose  upper  base  is  2"  in  diameter, 


lO  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

and  whose  height  is  2".    Let  the  pyramid  be  in  the  first  quadrant  with  its 
lower  base  on  H. 

4.  Draw  the  plan  and  elevation  of  a  \"  by  2"  by  3''  block  having  three 
circular  holes  through  it  the  \"  way.  The  holes  are  \"  in  diameter  and  are 
spaced  with  centers  \"  apart  with  the  middle  hole  in  the  center  of  the 
2"  by  4"  face  of  the  block. 

5.  Draw  the  plan  and  elevation  of  a  cube  with  2"  edges.  Let  the  cube 
be  in  the  first  quadrant  resting  on  H  with  its  vertical  faces  inclining  to  V  at 
angles  of  30  and  60  degrees. 

6.  Draw  the  plan  and  elevation  of  a  cast  iron  plate  9"  by  3'  by  6'.  In 
the  center  of  the  3'  by  6'  face  is  a  hole  d"  square  and  4"  deep;  in  the  9''  by  6' 
face  are  drilled  two  3"  holes  through  the  plate  8"  from  each  end  and  on  the 
center  fine. 

7.  A  grain  hopper  in  the  shape  of  a  regular  truncated  square  pyramid 
has  an  opening  4'  by  4',  and  an  outlet  12"  by  12"  which  is  4'  below  the 
opening.  Draw  the  plan  and  elevation  of  the  hopper  when  the  opening 
lies  in  the  H  plane  with  its  edges  inclining  at  angles  of  45  degrees  with  the 
G.  L. 

8.  Draw  the  plan  and  elevation  of  a  plain  box  ^'  deep,  6"  wide  and  9" 
long  with  its  lid  half  raised.  The  box  is  built  of  ^'  material  and  the  lid  is 
hinged  along  the  9"  edge. 

9.  A  hood  is  fastened  to  a  wall.  The  opening  of  the  hood  is  4'  by  4'  and 
its  outlet  is  into  a  square  pipe  6"  in  diameter  also  fastened  to  the  wall 
directly  above  the  opening.  The  distance  between  the  opening  and  outlet 
is  3'.    Draw  the  plan  and  elevation. 

10.  Draw  the  plan  and  elevation  of  an  hexagonal  pyramid  whose  base  is 
2"  in  diameter  and  whose  altitude  is  3".  The  apex  of  the  pyramid  is  directly 
above  one  corner  of  the  base. 

Note:  Unless  otherwise  specified  objects  given  in  problems  are  to  be 
drawn  in  the  third  quadrant;  and  when  no  distance  from  H  and  V  is  given 
the  views  may  be  located  at  any  convenient  distance  from  the  ground  fine. 
It  is  well,  however,  to  keep  the  distance  between  views  relatively  short  so 
that  the  eye  may  readily  note  which  points  are  projections  of  each  other; 
in  practical  drafting  only  enough  space  is  allowed  between  views  to  permit 
of  placing  dimensions  and  to  keep  the  views  distinct. 


CHAPTER   II 
PROFILE  PLANE 

11.  While  in  many  cases  a  plan  and  elevation  will  show  the 
shape  and  size  of  an  object  it  quite  frequently  happens  that  a 
third  view  will  be  required  for  a  complete  representation.  In 
such  a  case  the  H  and  V  planes  of  projection  are  not  sufficient 
and  a  third  plane  of  projection  must  be  added.  The  third  view, 
which  is  most  commonly 
used  to  supplement  the 
plan  and  elevation,  is 
called  an  efid  view,  or  a 
side  or  end  elevation,  or 
a  profile  projection;  and 
the  plane  upon  which 
this  view  is  made  is 
called  the  profile  plane, 
or  P  plane. 

12.  In  Fig.  8  is  shown 
a  picture  of  the  arrange- 
ment of  H,  V,  and  P 
for  the  third  quadrant. 
From  this  picture  it  will 
be  clear  that  the  profile 
plane  is  perpentiicular  to  the  G.  L.,  and  therefore  to  both  H 
and  V,  and  it  may  be  located  at  any  convenient  point  on  the 
G.  L.  and  on  either  or  both  sides  of  the  object.  In  case  it  is 
located  on  the  right  side  of  the  object  the  view  obtained  will  be 
the  right  end  view.  It  not  infrequently  occurs  in  drafting  that 
both  end  views  are  shown. 

13.  In  Fig.  9  is  shown  how  the  H,  V,  and  P  planes,  pic- 
torially  represented  in  Fig.  8,  appear  when  they  are  revolved  to 


12 


ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 


coincide  with  the  plane  of  the  drawing.  From  this  figure  it  will 
be  observed  that  the  relations  between  H  and  V  which  have  been 
previously  discussed  are  in  no  way  altered,  that  the  profile  plane 
takes  a  position  to  the  right  or  left  of  the  V  plane,  and  that  the 


HTPlane 


V.  Plane 


Elevation 


\ 


f--^. 


■\ 


End  View 


X 


/  p.  Plane 


Fig.  9. 


elevation  and  end  view  of  the  same  point  lie  on  lines  which  are 
parallel  to  the  G.  L.  It  will  also  be  observed  that  the  end  view 
shows  how  far  an  object  is  below  H,  by  means  of  the  distance  the 
end  view  is  below  the  G.  L.;  and  also  shows  how  far  an  object  is 
behind  V,  by  means  of  the  distance  the  end  view  is  to  the  right, 
or  left,  of  the  profile  ground  line. 

There  are,  of  course,  two  profile  ground  lines;  one  where  the 
P  plane  cuts  V,  and  the  other  where  it  cuts  H.  In  Fig.  9  it  will 
-be  noted  that  the  ground  line  between  V  and  P  is  really  an  end 


PROFILE  PLANE 


13 


view  of  the  V  plane,  and  the  one  between  H  and  P  is  really  an  end 
view  of  H.  This  latter  coincides,  after  the  planes  are  revolved, 
with  the  G,  L.  for  H  andjy:. 

In  actual  work  the  three  planes  are  not  limited  in  extent  and 
their  location  is  determined  only  by  the  ground  lines.    Fig.  10 


"~  '~~-,^ 

Plan 

"x 

\ 

1 
1 

\ 

\ 

' 

i         1 

I                        ! 

i      ■    1 

1 
1 

'  i 

-»  1 

1 
1 

1 

Elevation 

End  View 

Fig.  10. 


shows  how  Fig.  9  appears  as  usually  drawn  with  only  the  ground 
lines  showing  the  planes. 

14.  The  arrangement  of  the  three  planes  for  all  quadrants  is 
shown  pictorially  in  Fig.  11,  with  the  profile  plane  on  the  right. 
Now  it  will  be  observed  that  if  the  profile  plane  be  revolved  so 
that  the  portion  for  the  third  quadrant  fall  to  the  right  of  the 
V  plane  the  profile  plane  for  the  first  quadrant  will,  of  course, 
revolve  in  the  opposite  way,  and  will  be  coincident  with  V.    This 


14 


ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 


is  shown  in  Fig.  1 2  where  a  timber  is  partly  in  the  first  and  partly 
in  the  third  quadrant. 

In  practical  drafting,  however,  the  arrangement  of  the  three 
planes  for  the  first  quadrant  is  as  shown  in  Fig.  13  where  the 


Fig.  II. 


block  shown  at  A  is  projected.  The  P  plane  is  revolved  out  of 
the  way  of  the  elevation  in  the  same  way  as  for  the  third  quad- 
rant, thus  avoiding  the  confusion  which  might  result  from  having 
the  end  view  superimposed  upon  the  elevation. 

One  of  the  problems  constantly  before  the  draftsman  for  solu- 


PROFILE  PLANE 


15 


Fig.  12. 


s           \ 

Fig.  13. 

tion  is  the  construction  of  the  end  views  of  an  object  which  is 
shown  only  in  plan  and  elevation.  Fig.  14  shows  the  plan  and 
elevation  of  an  object  of  which  it  is  desired  to  find  both  end 
views. 

The  profile  ground  lines  are  assumed  at  any  convenient  point 
along  the  given  ground  line  as  shown.     Through  the  eleva- 


i6 


X 


ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 


tion  of  each  point  draw  dotted  lines  parallel  to  the  G.  L.  On 
these  dotted  lines,  the  proper  distance  from  the  profile  ground 
lines,  will  lie  the  end  view  of  each  point.     To  find  this  location 


Fig.  15. 

lay  off  on  the  dotted  lines  distances  to  the  right  and  left  of  the 
profile  ground  line  in  V,  the  distance  each  point  is  back  of  V. 
This  may  be  done  conveniently  by  drawing  a  45  degree  line  as 
shown  and  drawing  perpendiculars.     Since  the  distance  each 


PROFILE  PLANE 


17 


end  view  is  from  the  profile  ground  line  in  V  is  equal  to  the  dis- 
tance it  is  back  of  V,  points  so  located  will  be  the  proper  distance 
to  the  right  or  left  of  the  profile  G.  L. 

Fig.  15  shows  a  similar  problem  solved  in  the  first  quadrant 
and  serves  to  show  the  arrangement  of  views  for  the  first 
quadrant. 

By  using  similar  methods  of  projection  a  third  view  of  any 
object  may  be  obtained  when  two  views  are  given.  Thus :  a 
plan  view  may  be  found  from  the  elevation  and  end  view;  or  an 
elevation  may  be  found  from  the  end  view  and  plan. 


PROBLEMS   IN  PROJECTION 
II.  Draw  the  plan,  elevation,  and  right  end  view  of  a  cube  with  edges 
2"  long.    Let  the  upper  face  of  the  cube  be  parallel  to  and  i"  below  H  with 


-2^- 


n 


Fig.  16. 


Fig.  17. 


its  edges  inclining  30  and  60  degrees  to  the  G.  L.    Let  the  edge  of  the  cube 
nearest  V  be  \"  behind  V. 

12.  Draw  three  views  of  a  block  \"  by  2"  by  4".  In  the  center  of  one 
2"  by  4"  face  is  a  \"  square  hole  cut  half  way  through  the  block,  and  in  the 
center  of  each  \"  by  2"  face  is  a  hole  i"  in  diameter  drilled  \"  deep. 

13.  Draw  the  plan,  elevation,  and  left  end  view  of  a  bar  of  iron  4"  long, 


i8 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


2"  wide,  and  \"  thick  which  has  four  \"  holes  drilled  through  it  on  the  center 
line  of  the  2"  by  4"  face.  The  holes  are  spaced  with  their  centers  |"  apart. 
Let  the  face  of  the  bar  be  \"  below  H  and  parallel  to  it,  and  let  the  center 
line  of  this  same  face  incline  60  degrees  to  the  G.  L. 

14.  Draw  three  views  of  a  regular  truncated  hexagonal  pyramid.  Diam- 
eter of  large  base  3";  diameter  of  small  base  \\"\  distance  between  bases 
2".    Use  first  quadrant  projection  and  let  the  z"  base  be  in  V. 


*-w 


■^^ 


■^^ 


Fig.  18. 


-V/i- — »< IM- 


-^ 


-W* 


■m' 


1V4— 

h 

-^ 

-Drill  ^ 

.^ 

N 
r 

It 

Fig.  19. 


15.  A  hopper  has  an  opening  in  the  floor  4'  by  6'.  Its  12"  square  outlet 
is  5'  below  the  floor  with  one  edge  directly  under  the  center  of  one  6'  edge 
of  the  floor  opening.    Draw  three  views  of  the  hopper. 

16.  Draw  the  plan  and  elevation  and  find  the  right  end  view  of  the 
object  shown  in  Fig.  16. 

17.  Draw  the  plan  and  elevation  and  find  the  left  end  view  of  the  object 
shown  in  Fig.  17. 

18.  Draw  the  two  given  views  and  find  the  plan  of  the  object  shown  in 
Fig.  iS. 

19.  Draw  the  plan  of  the  object  whose  end  view  and  elevation  are  shown 
in  Fig.  19. 

20.  Draw  the  two  given  elevations  and  find  the  plan  of  the  object  shown 
in  Fig.  20. 


Fig.  20. 


Fig.  21. 


Fig.  22. 


(19) 


20 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


DriU  %" 


Fig.  23. 


Fig.  24. 

21.  Draw  three  views  of  the  object  shown  in  Fig.  21. 

22.  Draw  a  plan,  elevation,  and  left  end  view  of  the  object  shown  in 
Fig.  22. 

23.  Draw  the  plan,  elevation,  and  right  end  view  of  the  object  shown  in 
Fig.  23. 

24.  Draw  a  plan,  elevation,  and  both  end  views  of  the  object  shown  in 
Fig.  24. 


CHAPTER   III 


ASSUMPTION    OF   POINTS   AND    LINES 

15.  In  order  to  study  the  relation  of  various  points  and  lines 
which  collectively  form  the  representation  of  an  object,  and  to 
solve  problems  concerning  these  elements  of  the  representation, 
there  must  be  a  method  of  designating  points  and  lines,  and  the 
relation  that  these  points 
and  lines  have  to  the 
planes  of  projection  must 
be  known. 

In  drafting  a  point  is 
almost  never  designated 
by  a  letter,  but  in  descrip- 
tive geometry  it  has  been 
found  convenient  to  indi- 
cate points  and  lines  by 
this  means.  A  point  in 
space  is  indicated  by  the 
capital  letter,  as  A^  B,  C, 
etc.,  and  its  projections  are 
indicated  by  the  corre- 
sponding small  letter,  as 
a,  b,  c,  etc.;  a,  b,  c,  being 
used  to  indicate  the  plan 
view  of  the  point  and  a', 
b',  c',  etc.,  being  used 
to   indicate   its   elevation.  ^^'  ^^' 

Thus :  the  plan  view  of  the  point  A  is  a,  and  its  elevation  is  a', 
and  when  the  point  A  is  referred  to  it  means  the  actual  point 
in  space.  The  line  MN  means  a  line  in  space  whose  plan  view 
is  mn  and  whose  elevation  is  m'n'. 


/           \ 

(0                       \ 

V 

V 

f 

•// 

0 '         j  . 

22  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

The  relation  that  a  point  or  a  line  in  space  has  to  the  planes  of 
projection  may  be  discovered  from  its  projections.  In  Fig.  25 
is  shown  the  plan  and  elevation  of  a  hexagonal  pyramid  one  of 
whose  edges  is  OM.  Now  the  plan  view  of  the  point  0  is  o  and 
its  elevation  is  o',  and  an  inspection  of  the  figure  will  show  that 
the  distance  o  is  from  the  G.  L.  is  the  distance  that  O  is  from  the 
V  plane;  and  the  distance  o'  is  from  the  G.  L.  is  the  distance 
that  O  is  from  H.  It  is  also  to  be  seen  that  since  o  is  above  the 
G.  L.  the  point  in  space  is  behind  V,  and  since  o'  is  below 
the  G.  L.  the  point  in  space  is  below  H.  Therefore,  from  the 
drawing  it  may  be  found  that  the  point  0  Hes  in  the  third  quad- 
rant, and  its  distances  from  H  and  V  may  be  measured.  In  Hke 
manner  the  relation  that  M  bears  to  H  and  V  may  be  read  from 
the  drawing. 

16.  To  assume  a  point  in  space  it  is  necessary  only  to  assume 
its  two  projections  for,  in  general,  a  point  in  space  is  completely 
located  with  reference  to^H  and  V  when  its  plan  and  elevation 
are  shown.  If,  however,  the  problem  is  to  assume  a  point  in  a 
given  quadrant  at  a  given  distance  from  H  and  V,  the  two  views 
must  be  located  at  the  proper  distances  from  the  G.  L.  Thus, 
in  Fig.  25,  0  which  is  to  he  in  the  third  quadrant  i"  from  H  and 
2"  from  V  will  be  drawn  as  follows:  Erect  the  dotted  line  00' 
perpendicular  to  the  G.  L.;  on  this  Hne  will  lie  o  and  o',  the  plan 
and  elevation  of  the  required  point.  Since  this  required  point 
is  the  third  quadrant  it  will  be  below  H  and  behind  V;  therefore, 
the  plan  view  o  will  be  above  the  G.  L.  a  distance  equal  to  the 
distance  the  point  is  behind  V,  or  2" .  Therefore,  locate  o  2" 
above  the  G.  L.  on  the  dotted  perpendicular.  Since  the  point 
is  \"  below  H,  o'  will  lie  in  this  same  perpendicular  \"  below  the 
G.  L. 

In  like  manner  a  point  may  be  located  in  any  of  the  quadrants 
and  at  any  specified  distance  from  H  and  V. 

17.  To  assume  a  line  in  space  assume  its  plan  and  elevation 
in  any  desired  position.  If  two  points  of  the  line  are  given,  the 
plan  and  elevation  of  the  line  may  be  found  by  joining  the  plan 
views  of  the  two  given  points,  and  the  elevation  by  joining  the 
elevations  of  the  two  given  points.     The  plans  and  elevations 


ASSUMPTION  OF  POINTS  AND  LINES 


23 


of  the  given  points  will  have  to  be  located  as  indicated  in 
Article  16. 

If  the  plan  and  elevation  of  a  line  are  given  and  it  is  desired 
to  assume  a  point  on  the  line  it  is  necessary  only  to  assume  the 
plan  view  of  the  point  on  the  plan  view  of  the  line  and  the  eleva- 
tion of  the  point  on  the  elevation  of  the  line,  and  have  these  two 
views  lie  in  the  same  per- 
pendicular to  the  G.  L. 

18.  To  assume  two  in- 
tersecting lines  assume 
two  lines  which  have  a 
common  point.  Fig.  26 
shows  AB  and  BC  inter- 
secting at  the  common 
point  B. 

19.  There  are  certain 
relations  which  points 
and  lines  have  to  the 
planes  of  projection  which 
are  apparent  from  a  study 
of  the  foregoing  articles 
and  these  relations  may 
be  set  forth  as,  — 

Observations. 

a.  The  two  views  of  a 
point  always  lie  in  a  per- 
pendicular to  the  G.  L. 

h.  The  distance  a  point 
is  from  H  is  equal  to  the  ^^^'  ^^• 

distance  its  elevation  is  from  the  G.  L.;    and  the  distance  a 
point  is  from  V  is  equal  to  the  distance  its  plan  is  from  the  G.  L. 

c.  A  point  which  hes  in  H  will  coincide  with  its  own  plan  view 
and  its  elevation  will  be  in  the  G.  L.  Likewise  a  point  in  V  will 
coincide  with  its  own  elevation  and  its  plan  view  will  He  in  the 
G.  L. 

d.  A  line  which  lies  in  H  or  V  will  coincide  with  its  view  on  H 
or  V  and  will  have  its  other  view  in  the  G.  L. 


24 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


,  e.  When  a  Une  is  parallel  to  H  or  V  its  view  on  H  or  V  will  be 
equal  in  length  to  the  Une  and  its  other  view  will  be  parallel  to 
the  G.  L. 


A  convenient  method  of  showing  the  relation  of  the  H,  V,  and  P  planes  for  the 
third  quadrant,  and  of  the  projections  of  any  given  line  on  the  planes. 


A  convenient  method  of  visualizing  any  given  problem  in  projection.  The  pencil 
is  held  in  the  position  of  the  given  line  and  its  projections  on  the  cardboard  planes 
may  be  drawn. 

/.  When  a  line  is  perpendicular  to  H  or  V  its  view  on  H  or  V 
will  be  a  point.  The  other  view  will  be  equal  in  length  to  the 
line  itself  and  perpendicular  to  the  G.  L. 


ASSUMPTION  OF  POINTS  AND   LINES  25 

g.  When  a  line  is  parallel  to  the  G.  L.  both  views  of  the  line 
will  be  parallel  to  the  G.  L. 

h.  The  projection  of  a  hne  on  H  or  V  will  be  equal  in  length  to 
the  line  itself  when  the  line  is  parallel  to  H  or  V. 

i.  The  projection  of  a  line  on  H  or  V  is  either  equal  in  length 
to  the  line  or  shorter  than  the  line. 

j.  A  Hne  may  lie  in  one  or  in  two  or  three  quadrants,  but  never 
in  all  four. 

k.  If  a  point  is  on  a  line  the  plan  view  of  the  point  will  He  on 
the  plan  view  of  the  Hne,  and  the  elevation  of  the  point  will  He 
on  the  elevation  of  the  Hne. 

/.  When  a  Hne  is  paraHel  to  H  and  oblique  to  V  the  elevation 
of  the  Hne  is  parallel  to  the  G.  L.  and  the  plan  view  makes  an 
angle  with  the  G.  L.  equal  to  the  angle  the  Hne  makes  with  V. 

In  like  manner  when  the  Hne  is  parallel  to  V  and  incHned  to  H 
its  plan  is  parallel  to  the  G.  L.  and  its  elevation  makes  an  angle 
with  the  G.  L.  equal  to  the  angle  the  line  makes  with  H. 

m.  When  two  Hues  intersect  they  have  a  point  in  common. 
Therefore,  the  plan  views  of  these  lines  will  have  a  point  in  com- 
mon, and  the  elevations  will  have  a  point  in  common. 

n.  Two  Hues  which  are  parallel  to  each  other  in  space  will  have 
plan  views  and  elevations  which  are  respectively  parallel. 

20.  These  observations  should  be  verified  by  experiment.  A 
convenient  method  of  doing  this  which  at  once  serves  to  verify 
the  experiment  and  at  the  same  time  to  assist  the  experimenter 
to  visuaHze  the  problem  is  shown  in  the  illustrations.  Two 
pieces  of  cardboard  are  cut  to  form  the  planes  of  projection  and 
a  pencil  or  a  hat  pin  —  which  is  easier  to  secure  in  place  —  i^ 
held  in  the  position  the  Hne  occupies  in  space.  The  projections 
of  this  position  may  be  plotted  on  the  cardboard  and  the  planes 
revolved,  thus  showing  the  relation  of  the  plan  and  elevation  of 
the  position  to  the  G.  L.  These  positions  should,  of  course, 
conform  to  the  positions  indicated  in  the  observations. 


26  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

PROBLEMS  IN  LOCATION  OF  POINTS  AND  LINES 

25.  Locate  a  point  M  i"  behind  V  and  2''  below  H. 

26.  Locate  a  point  N  Y'  in  front  of  V  and  i^"  above  H. 

27.  Locate  a  point  O  i"  behind  V  and  i|"  above  H. 

28.  Locate  a  point  P  2"  in  front  of  V  and  i"  below  H. 

29.  Locate  a  point  M  in  the  third  quadrant  2"  from  H  and  i''  from  V. 

30.  Locate  a  point  N  in  the  first  quadrant  i^"  from  H  and  2"  from  V. 

31.  Locate  a  point  O  in  the  second  quadrant  2"  from  H  and  V. 

32.  Locate  a  point  P  in  the  fourth  quadrant  i"  from  H  and  2"  from  V. 

33.  Assume  a  line  MN  in  the  third  quadrant  parallel  to  H  and  i"  below 
H,  and  making  an  angle  of  30  degrees  with  V. 

34.  Assume  a  line  in  the  first  quadrant  parallel  to  V  and  2"  from  it  and 
inclining  to  H  at  60  degrees. 

35.  Assume  a  Hne  in  the  third  quadrant  parallel  to  V  and  2"  from  V, 
and  making  an  angle  of  30  degrees  with  H. 

36.  Draw  a  line  in  the  first  quadrant  parallel  to  the  G.  L.  and  i  ^"  from 
H  and  V. 

37.  Assume  a  line  with  one  end  in  the  first  and  the  other  end  in  the  third 
quadrant.    Draw  an  end  view  of  this  line. 

38.  Draw  a  line  parallel  to  H  and  making  an  angle  of  30  degrees  to  V  in 
both  first  and  second  quadrants. 

39.  Assume  a  point  in  the  third  quadrant  2''  from  H  and  i"  from  V. 
Through  this  point  draw  two  intersecting  lines,  one  parallel  to  H  and  one 
parallel  to  V. 

40.  Assume  two  intersecting  lines  in  the  first  quadrant,  one  parallel  to 
H  and  one  parallel  to  V  and  make  them  3"  long.  Let  the  point  of  inter- 
section be  2"  from  H  and  Y'  from  V. 

41.  Assume  a  line  3"  long  in  the  third  quadrant  parallel  to  the  G.  L.  and 
i"  from  H  and  V.  Through  the  center  of  this  line  draw  another  line  making 
an  angle  of  45  degrees  with  both  H  and  V.  Draw  an  end  view  of  the  two 
lines. 

42.  One  line  which  inclines  45  degrees  to  both  H  and  V  is  intersected  by 
another  line  which  inclines  30  degrees  to  H  and  60  degrees  to  V.  The  point 
of  intersection  is  2"  from  H  and  V.  Draw  three  views  of  the  two  lines  in 
the  third  quadrant. 

43.  Draw  a  hne  parallel  to  V  i"  behind  it,  and  making  60  degrees  with 
H.    Draw  a  second  line  parallel  to  this  line  i  h"  from  it. 

44.  A  hne  is  2"  below  H  and  parallel  to  H.  One  end  is  in  V  and  the 
other  end  is  3"  behind  V,  and  between  these  points  the  line  is  4"  long. 
Through  the  middle  point  of  this  Hne  draw  a  second  line  parallel  to  V. 


CHAPTER   IV 
PLANES 

21.  A  plane  is  represented  in  descriptive  geometry  by  its 
traces.  These  traces  are  the  lines  in  which  the  plane  intersects 
H,  V,  and  P,  and  are  designated  as  the  H  trace,  the  V  trace,  and 
the  P  trace. 

CSince  the  G.  L.,  which  is  common  to  both  H  and  V,  can  meet 
the  plane  in  but  one  point  it  will  be  obvious  that  the  H  trace 
and  the  V  trace  must  meet  each  other  at  this  point.  In  designat- 
ing planes  on  a  drawing  this  point  is  marked  S,  T  or  R,  etc., 
while  the  H  trace  is  marked  with  the  corresponding  small  letter 
s,  t,  or  r,  etc. ;  and  the  V  trace  s',  t',  or  r',  etc. ;  and  the  P  trace 
Sp,  tp,  or  rp,  etc. 

In  drafting,  planes  are  usually  so  arranged  as  a  part  of  the 
object  being  drawn  that  they  are  represented  by  limiting  lines. 
For  example  in  a  cube  the  faces  are  shown  in  plan  and  elevation 
as  squares  and  the  traces  of  the  planes  of  the  faces  are  not  needed. 
But  while  the  planes  of  any  face  of  an  object  are  usually  not 
needed  to  make  the  complete  representation,  yet  they  are  useful 
in  solving  drawing-board  problems  which  arise  in  connection 
with  the  drawing  of  the  objects.  They  should,  therefore,  be 
studied  with  the  idea  in  mind  that  they  are  a  means  to  an  end 
rather  than  an  end  in  themselves.  For  this  reason  it  will  be 
convenient  to  represent  traces  with  a  dashed  line  arranged  as 
shown  in  the  figures  following. 

22.  Planes  in  descriptive  geometry  are  considered  to  be 
indefinite  in  extent.  For  this  reason,  then,  it  should  be  clear 
that  any  plane  extends  through  all  four  quadrants  even  though 
its  traces  show  it  as  being  in  only  one.  It  is  customary  to  show 
the  traces  of  a  plane  only  in  that  quadrant  where  the  problem 
lies,  yet  if  it  were  necessary  these  traces  could  be  extended. 

27 


28  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

In  Fig.  27  is  shown  a  plane  in  the  first  quadrant;  in  Fig.  28  a 
plane  in  the  second  quadrant;  in  Fig.  29  a  plane  in  the  third 
quadrant;  and  in  Fig.  30  a  plane  in  the  fourth  quadrant.     Fig. 


/     //     / 
/      //      / 

< f ^ 


A 


^        7V~" 
■■  \       /        \ 

t 

Fig.  27.  Fig.  28.  Fig.  29.  Fig.  30. 


/ 


t 

Fig.  31. 


31  shows  a  plane  indefinite  in  extent  and  therefore  in  all  four 
quadrants:  it  is  obvious  from  this  figure  that  any  pair  of  the 
given  H  or  V  traces  will  locate  this  same  plane. 


PLANES  29 

23.  Observations. 

a.  The  traces  of  a  plane  which  is  parallel  to  the  G.  L.  will 
themselves  be  parallel  to  the  G.  L.     Fig.  32  shows  such  a  plane. 

h.  If  two  planes  are  parallel  their  respective  traces  will  be 
parallel. 

c.  If  a  plane  is  perpendicular  to  H,  its  V  trace  will  be  perpen- 
dicular to  the  G.  L.  Likewise,  when  it  is  perpendicular  to  V  its 
H  trace  will  be  perpendicular  to  the  G.  L. 

S' * 


Fig.  32. 


d.  If  a  plane  is  perpendicular  to  H  it  will  also  be  perpendicular 
to  any  other  plane  whose  H  trace  is  perpendicular  to  its  H  trace. 
Likewise  a  plane  perpendicular  to  V  will  be  perpendicular  to  any 
other  plane  whose  V  trace  is  perpendicular  to  its  V  trace. 

e.  If  the  plan  and  elevation  of  a  line  are  perpendicular  respec- 
tively to  the  H  and  V  traces  of  a  plane  the  line  will  be  perpendicu- 
lar to  the  plane. 

/.  If  a  Une  is  perpendicular  to  a  plane,  the  plan  view  of  the 
line  will  be  perpendicular  to  the  H  trace  of  the  plane  and  the 
elevation  of  the  line  will  be  perpendicular  to  the  V  trace  of 
the  plane. 

g,  A  plane  is  determined  by  any  two  intersecting  Hnes,  any 
three  points,  any  two  parallel  lines,  or  a  point  and  Une.  The 
plane  is,  of  course,  in  each  case  represented  by  its  traces. ' 

h.  The  plan  view  of  the  V  trace  of  any  plane  is  in  the  G.  L. ; 

and  the  elevation  of  the  H  trace  of  any  plane  is  also  in  the  G.  L. 

">^  i.  If  a  line  lies  in  a  plane  and  is  parallel  to  the  H  trace  its  plan 

view  will  be  parallel  to  that  trace  and  its  elevation  will  be  parallel 


30 


ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 


to  the  G.  L.  Likewise,  when  the  line  is  parallel  to  the  V  trace 
its  elevation  will  be  parallel  to  that  trace  and  its  plan  will  be 
parallel  to  the  G.  L. 

These  observations  are  collected  here  for  convenience  and 
reference.  No  attempt  has  been  made  to  prove  the  truth  of  the 
assertions  set  forth  because  they  are  more  or  less  axiomatic,  and 
both  the  proof  and  truth  of  these  facts  will  be  obvious  as  the 

study  of  the  subject 
progresses,  and  may  be 
proved  by  experiment 
as  in  Article  19. 

24.  To  assume  a 
plane,  assume  any  two 
traces  which  are  parallel 
to  the  G.  L.,  or  which 
intersect  the  G.  L.  at  a 
common  point. 

25.  To  assume  a  line 
in  a  plane,  assume  either 
view  of  the  line,  as  the 
plan  view  ab  in  Fig.  33. 
The  plan  view  when  ex- 
tended cuts  the  G.  L.  at 
m  and  cuts  the  H  trace 
of  the  plane  at  n.  Since 
n  is  on  the  H  trace  its 
elevation  will  be  on  the 
G.  L.  at  n';  and  since  m 
is  on  the  G.  L.,  m'  — 


Fig.  33. 


since  the  line  AB  is  to  be  in  the  plane  T  —  must  lie  on  the  V 
trace,  m'n',  therefore,  is  the  elevation  and  mn  the  plan  of  MN 
lying  in  the  plane  T.  ab  and  a'b'  are,  of  course,  on  mn  and 
m'n'  and  are  the  plan  and  elevation  of  a  line  AB  lying  in  the 
plane  T. 

26.  To  assume  a  point  in  a  plane. — One  method  is  to  assume 
a  line  in  the  plane  —  as  in  the  foregoing  article  —  and  on  this 
line  assume  the  point.    Another  method  more  convenient  to  use 


PLANES 


31 


is  to  assume  one  view  of  the  point  —  as  the  plan  view  a  in  Fig. 
34  —  and  through  this  assumed  plan  view  of  the  point  draw  the 
plan  mn  of  a  line  MN  parallel  to  the  H  trace.  Since  this  line 
MN  is  parallel  to  the  H  trace  its  elevation  will  be  parallel  to  the 


Fig.  34. 

G.  L. ;  mn  and  m'n',  then,  are  the  plan  and  elevation  of  a  line 
MN  lying  in  the  plane  T.  Since  a  Hes  on  mn,  a'  will  lie  on 
m'n',  and  the  point  A  then,  as  represented  by  the  plan  a  and 
elevation  a',  will  be  a  point  in  the  plane  T. 


PROBLEMS   ON  PLANES 

45.  Assume  a  plane  parallel  to  the  G.  L.  so  that  its  H  trace  is  2"  back 
of  V  and  its  V  trace  i"  above  H.     Show  an  end  view  of  this  plane. 

46.  A  plane  is  perpendicular  to  H  and  inclined  30  degrees  to  V.     Show 
its  three  traces. 

47.  Draw  a  plane  parallel  to  the  plane  in  Problem  46.    Draw  the  profile 
traces  2"  apart. 


32  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

48.  Draw  the  H,  V,  and  P  traces  of  a  plane  in  the  third  quadrant.  Let 
the  H  trace  inchne  60  degrees  to  the  G.  L.  and  the  V  trace  45  degrees  to  the 
G.  L. 

49.  Assume  any  line  in  the  first  quadrant  obUque  to  H  and  V  and  show 
the  traces  of  a  plane  perpendicular  to  this  line. 

50.  Assume  any  oblique  plane  in  the  third  quadrant  and  show  the  traces 
of  a  plane  which  will  be  perpendicular  to  this  plane  and  to  V. 

.  51.  Assume  a  plane  whose  H  and  V  traces  incline  60  degrees  to  the  G.  L. 
In  this  plane  is  cut  a  hole  whose  plan  view  is  a  i"  square.  Assume  the  loca- 
tion of  the  plan  view  of  the  hole  and  show  its  plan,  elevation,  and  end  view. 

52.  Assume  a  plane  whose  H  trace  makes  45  degrees  with  the  G.  L.  and 
whose  V  trace  makes  60  degrees  with  the  G.  L.  In  this  plane  is  cut  a  hole 
whose  elevation  is  a  circle  i"  in  diameter.  Find  also  the  plan  and  end  view 
of  this  hole.    The  location  of  the  elevation  may  be  assumed. 

53.  A  plane  inclines  60  degrees  to  V  and  30  degrees  to  H  and  is  parallel 
to  the  G.  L.  3"  from  it.  In  this  plane  is  cut  a  hole  i'^  square;  show  its  plan, 
elevation,  and  end  view. 

54.  A  plane  inclines  60  degrees  to  V  and  is  perpendicular  to  H.  In  this 
plane  is  cut  a  i"  circular  hole  with  its  center  1^"  below  H.  Show  three 
views  of  the  plane  with  the  hole  cut  in  it. 


CHAPTER  V 

LOCATION    OF   POINTS,   LINES,   AND    PLANES 

27.  Character  of  Lines.  In  order  to  facilitate  the  reading  of 
the  drawings  in  the  text  all  given  and  required  Hnes  will  be  drawn 
heavy  and  all  auxiliary  and  constructive  lines  will  be  drawn  light. 
When  necessary  the  two  views  of  a  point  will  be  connected  by  a 
light  dotted  line  perpendicular  to  the  ground  Une,  but  wherever 
possible  the  hne  is  omitted  to  avoid  the  confusion  of  unnecessary 
lines.     The  following  chart  shows  the  character  and  purpose  of 


Visible  outUne 


Invisible  outline 


Trace  of  plane 
Path  of  moving  point 
Joining  projections  of  point 
Center  line 


CHART  OF  LINES 

each  line  used  in  the  drawings  and  will  be  found  convenient  for 
reference. 

28.  Method  of  Locating  Points  and  Planes.  —  A  point  will  be 
located  by  giving  its  distances,  or  coordinates,  in  inches  always 
unless  otherwise  marked,  from  P,  V,  and  H  and  in  that  order. 
Thus:  A  is  o;  2;  2  means  that  A  in  space  is  in  the  profile  plane, 
two  inches  from  V,  and  two  inches  from  H;  or  B  is  3 ;  4;  i  means 
that  B  is  three  inches  from  the  profile  plane,  four  inches  from  V, 
and  one  inch  from  H. 

33 


34 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


In  order  to  indicate  on  which  side  of  the  planes  of  projection  a 
point  lies,  —  or,  in  other  words,  in  which  quadrant  it  is,  —  plus 
and  minus  signs  will  be  used.  Plus  always  means  to  the  right 
of  P,  in  front  of  V,  and  above  H,  and  unless  otherwise  marked 
points  will  be  understood  to  be  plus.  Thus  A  is  4;  3;  2  means 
that  A  is  four  inches  to  the  right  of  the  profile  plane,  three  inches 
in  front  of  V,  and  two  inches  above  H.  In  other  words,  it  is  in 
the  first  quadrant. 

Minus  always  means  to  the  left  of  P,  behind  V  and  below  H. 
Thus:  B  is  —2;  —3;  —5  means  that  B  is  two  inches  to  the 
left  of  P,  three  inches  behind  V,  and  five  inches  below  H.  B,  in 
other  words,  is  in  the  third  quadrant.  It  will  be  convenient  to 
remember  that  a  plus  sign  preceding  the  last  two  figures  means 
that  the  point  is  in  the  first  quadrant.  Thus  M  is  —4;  +2;  +3 
means  M  is  in  the  first  quadrant  four  inches  to  the  left  of  P,  two 
inches  in  front  of  V,  and  three  inches  above  H.  Nis2;— i;— 2 
means  that  N  is  in  the  third  quadrant  two  inches  to  the  right  of 
P,  one  inch  behind  V,  and  two  inches  below  H. 

A  line  will  be  located  by  giving  reference  points  for  any  two  of 
its  points  but  it  is  to  be  understood  that  these  two  points  do  not 
necessarily  limit  the  extent  of  the  line.  Thus  Mis— 2;  — 3;  — 4; 
N  is  3;  —4;  —2  means  that  the  line  MN  is  in  the  location  indi- 
cated by  the  location  of  M  and  N,  but  it  is  not  necessarily  limited 
in  length  by  M  and  N. 

A  plane  will  be  located  by  giving  the  distance  its  point  on  the 
ground  line  is  from  P,  and  next  the  angles  its  traces  make  with 
the  ground  line.  Thus  T  is  —4;  30;  45  means  that  T  is  so 
located  that  the  point  where  its  traces  cut  the  G.  L.  is  four  inches 
to  the  left  of  P  and  that  the  plane  slopes  so  that  its  V  trace  in- 
,  clines  30  degrees  to  the  G.  L.  and  its  H  trace  slopes  45  degrees 
to  the  G.  L.  The  intersection  of  the  traces  on  the  G.  L.  is  the 
vertex  of  the  angle,  and  to  indicate  the  direction  of  the  traces 
the  angles  that  V  traces  make  above  the  G.  L.  on  the  right  of  this 
vertex  are  marked  minus,  and  the  angles  H  traces  make  on  the 
right  of  this  vertex  below  G.  L.  are  plus.  Thus  T  is  o;  -45;  45 
means  that  the  point  T  is  in  the  profile  plane,  that  the  V  trace 
makes  an  angle  of  45  degrees  with  the  G.  L.,    and  is  above  the 


LOCATION  OF  POINTS,  LINES,   AND   PLANES  35 

G.  L.  and  to  the  right  of  T,  and  that  the  H  trace  is  below  the 
G.  L.  and  makes  an  angle  of  45  degrees  with  it  on  the  right  of  T. 
S  is  —2;  45;  —45  means  that  the  point  S  on  the  G.  L.  is  two 
inches  to  the  left  of  the  profile  plane,  that  the  V  trace  is  below 
the  G.  L.  and  makes  an  angle  of  forty-five  degrees  with  it  on 
the  right  of  the  point  T,  and  the  H  trace  is  above  the  G.  L.  and 
makes  an  angle  of  forty-five  degrees  with  it  on  the  right  of  the 
point  T. 

When  a  plane  is  parallel  to  the  ground  line  there  is,  of  course, 
no  point  on  the  G.  L.  and  the  traces  make  no  angle  with  the  G.  L. 
Such  a  plane  is  indicated  thus :  00  ;  314  which  means  that  the 
point  S  is  at  infinity  —  thus  making  the  traces  parallel  to  the 
G.  L.  —  and  that  the  V  trace  is  three  inches  above  the  G.  L.  and 
the  H  trace  is  four  inches  below  the  G.  L.  A  plane  marked 
00  ;  —3;  —4  means  that  the  traces  are  parallel  to  the  G.  L.,  the 
V  trace  being  three  inches  below  the  G.  L.  and  the  H  trace  being 
four  inches  above  the  G.  L. 

29.  Accuracy  and  Checking.  —  In  working  the  problems  on 
the  drawing  board  it  is  reasonable  to  expect  results  ^thin  two 
hundredths  of  an  inch.  That  is,  if  a  line  is  actually  two 
inches  long  its  length  should  be  found  within  the  limits  one 
and  ninety-eight  hundredths  and  two  and  two  hundredths 
inches.  Such  a  degree  of  accuracy  means,  of  course,  great 
care  in  drawing  and  plotting.  An  error  made  at  the  beginning 
of  a  problem  is  apt  to  accumulate  as  the  problem  progresses, 
and  the  only  way  to  get  close  results  is  to  check  the  problem 
step  by  step. 

Checking  means  verifying  a  result  by  another  method,  and  it 
is  important  in  drafting  operations  that  this  means  of  verifying 
work  be  resorted  to  often. 

PROBLEMS  ON  THE  LOCATION  OF  POINTS,  LINES,  AND   PLANES 

55.  A  point  is  in  the  third  quadrant  3"  to  the  left  of  the  profile  plane,  2" 
from  H  and  4"  from  V.  Write  these  distances  in  the  proper  order,  and  with 
the  correct  signs.     Show  also  three  views  of  the  point. 

56.  The  point  m  is  3"  above  H,  2"  in  front  of  V,  and  in  the  profile  plane. 
Show  three  views  and  give  the  coordinates  of  the  point. 


36  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

57.  Locate  the  following  points  and  indicate  in  what  quadrant  each  is: 

M  =  o;  2";  1" 

58.  M  =  o;  -2";  -1".  N  =  +2;  -i'';  -\".  O  =  1")  -2"',  -2". 
From  O  draw  a  line  OP  to  the  middle  point  of  the  line  MN  and  give  its 
coordinates. 

59.  A  =  -3";  +2-;  +1".  B  =  -i'';  +4";  +2".  C  =  -T;  +1"; 
+  1".    Draw  the  plan,  elevation,  and  end  view  of  the  triangle  ABC. 

60.  M  =  +2;— i;— I.  N=  — 2;-f-2;H-2.  Draw  a  plan,  elevation, 
and  end  view  of  this  line.     Use  a  profile  plane  at  —3  for  the  end  view. 

61.  A  =  o;  —2;  +1.  B  =  o;  +2;  —  i.  Show  a  plan,  elevation,  and 
end  view  of  this  hne. 

62.  M  X  o;  — 3;  o.  N  =  o;  o;  —3.  Show  a  plan,  elevation,  and  end 
view  of  this  Hne  and  from  the  G.  L.  erect  a  perpendicular  to  it.  Give  the 
coordinates  of  the  point  of  intersection. 

63.  A  =  —3;  —2;  — i;  B  =  — i;  —2;  —4.  With  A  and  B  as  one  side 
construct  a  square  parallel  to  V  and  show  three  views  of  it. 

64.  M  =  o;  —3;  —2;  N  =  +3;  o;  +2;  0  =  +1;  —  i;  o.  Draw  three 
views  of  the  triangle  MNO. 

65.  Show  three  views  of  the  solid  —  one  of  whose  faces  is  ABCD  — 
whose  corners  are  given  below: 

A  =  o;— 2;— 2;B=  +  2;-2;— 2;C  =  +2;— 2;  -4;  D  =  o;  —  2;  -4; 
E  =  o;  -4;  -4;  F  =  o;  -4;  -2;  G  =  +2;  -4;  -2;  H  =  +2;  -4;  -4. 

66.  A  =  +2;  -z\\  -2;  B  =  +2;  -\\\  -  2;  C  =  +2;  -  2\\  -  3J; 
0  =  -|-6;  — 2j;  — 2f.  Draw  three  views  of  this  pyramid  considering  ABC 
as  the  base  and  O  the  apex. 

67.  A  cube  2"  on  each  edge  has  one  corner  at  +2";  —3";  o.  The  edge, 
of  which  this  point  is  one  end,  slopes  away  from  H  at  an  angle  of  30  degrees 
to  the  left  and  is  parallel  to  V.  Draw  three  views  of  the  cube  in  the  third 
quadrant. 

68.  Locate  the  three  traces  of  the  plane  o;  —30;  +60. 

69.  Locate  the  three  traces  of  the  plane  —2;  +60;  —45. 

70.  Locate  the  three  traces  of  the  plane  +1;  +60;  +30. 

71.  Locate  the  three  traces  of  the  plane  o;  —45;  —75. 

72.  Locate  the  three  traces  of  the  plane  o;  — 135;  +45. ' 

73.  Locate  the  three  traces  of  the  plane  o;  —150;  +105. 

74.  Locate  the  three  traces  of  the  plane  —  i;  —165;  +150. 

75.  Locate  the  three  traces  of  the  plane  +2;  —90;  -I-90. 

76.  Locate  the  three  traces  of  the  plane  —2;  +90;  —75. 

77.  Locate  the  three  traces  of  the  plane  —3; +135;— 135. 


LOCATION  OF  POINTS,   LINES,  AND   PLANES  37 

78.  The  plane  S  =  o;  —45;  +60.    Assume  a  line  in  this  plane  2"  above 
and  parallel  to  H. 

79.  The  plane  T  =  —2;  —135;  +120.    Assume  a  line  in  this  plane  \" 
behind  and  parallel  to  V. 

80.  The  plane  S  =  o;  +60;  —60.    Assume  a  line  in  the  plane  2"  below 
H  and  parallel  to  H. 

81.  The  plane  s  =  —2";  +120;  —135.    Assxmie  a  line  in  this  plane 
parallel  to  V  and  \"  behind  V. 

82.  Locate  the  planes 

<»;  -3;  -4 

00;  -3;  00 

00  ;  00  ;  -f  3 
00;  00  ;  —2 

^\-^2\  -I. 

83.  The  plane  S  =  00  ;  +3";  —4".    Find  the  coordinates  of  a  line  3" 
long  in  this  plane  2"  from  the  G.  L. 

84.  The  plane  T  =  oo  ;  +2";  —3".    Draw  a  hole  1"  square  in  this  plane, 
and  show  three  views  of  it. 


CHAPTER   VI 

REVOLUTION    OF   POINTS 

30.  One  of  the  most  important  operations  in  descriptive 
geometry  and  the  means  by  which  a  great  many  problems  are 
solved  is  the  revolution  of  a  point,  or  a  line,  from  one  position  to 
another.  The  actual  laws  governing  such  a  movement  are  well 
known,  but  to  represent  the  change  of  position  correctly,  and  to 
visualize,  or  to  get  a  mental  picture  of  what  is  taking  place  dur- 
ing the  change  of  position,  requires  careful  study. 


Fig.  35.    . 

31.  When  a  point  revolves  about  a  line  two  things  are  ap- 
parent at  once: 

1.  That  the  point  describes  a  circle  in  space. 

2.  That  the  plane  of  this  circle  is  perpendicular  to  the  line. 
Put  in  other  words,  the  revolution  of  any  point  about  an  axis 

takes  place  in  a  plane  perpendicular  to  the  axis. 

32.  Fig.  35  shows  the  plan  and  elevation  of  a  circle  which 
has  been  made  by  revolving  the  point  A  about  the  line  MN  which 
Kes  in  H.    From  this  figure  it  will  be  seen  that  when  A  lies  in  H, 

38 


REVOLUTION  OF  POINTS 


39 


or  the  plane  of  the  axis,  it  will  be  at  Ah  and  that  the  distance  from 
Ah  to  the  axis  is  equal  to  the  radius  of  the  circle. 

In  terms  of  the  distances  which  are  shown  by  the  plan  and 
elevation  of  point  A,  this 
radius  is  equal  to  the  hypot- 
enuse of  a  right  triangle 
whose  base  is  equal  to  the 
distance  a  is  from  the  plan 
of  the  axis,  or  ax,  and  whose 
altitude  is  equal  to  the  dis- 
tance a'  is  below  the  G.  L., 
or  the  distance  A  is  below 
the  plane  of  the  axis,  or  a'g. 
Since  the  plane  of  revolu- 
tion is  perpendicular  to  the 

axis,  A  will  fall  on  a  per-  "  \, ^"^ 

pendicular  through  a  to  mn,  ^^^'  36- 

either  on  the  right  or  the  left  of  MN,  a  distance  from 
MN  equal  to  a'n',    or   the   hypotenuse   of   a   triangle    whose 

base  is  ax  and  whose  altitude 
is  a'g. 

33.  Fig.  36  shows  this  same 
problem  worked  out  when 
the  axis  MN  lies  in  H  but 
oblique  to  V.  In  this  case 
Ah  will  be  found  on  a  per- 
pendicular from  a  to  mn 
whose  length  is  equal  to  the 
hypotenuse  of  a  triangle  whose 
base  is  ao,  and  whose  altitude 
is  the  distance  from  a'  to  the 
G.  L.,  or  a'g. 
Fig.  37  shows  a  similar 
■  ^'^'  problem  worked  out  with  the 

point  revolved  into  V  about  an  axis  which  Hes  in  V.  In  this 
case  the  point  A  falls  on  a  perpendicular  to  m'n'  whose  length  is 
equal  to  the  hypotenuse  of  a  triangle  whose  altitude  is  equal  to 


40 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


the  distance  a  is  from  the  G.  L.,  and  whose  base  is  the  dis- 


tance a'o'. 


Fig.  38. 

34.  A  line  may  be  revolved  into  H  or  V  about  an  axis  which 
lies  in  H  or  V  by  revolving  any  two  of  its  points.     The  line  and 

axis  must,  however,  lie  in 
the  same  plane  else  one 
point  of  the  revolving  line 
would  touch  H  or  V  before 
the  other.  Fig.  38  shows  a 
line  AB  revolved  into  H  at 
ArBh  about  MN  which 
lies  in  H  and  is  parallel  to 
AB.  The  location  of  the 
points  Ah  and  Bh  may 
be  found  as  indicated  in 
Article  32. 

35.  Either  a  point  or  a 
line  may  be  revolved  about 
an  axis  into  the  plane  of  the  axis  when  the  plane  does  not  coin- 
cide with  the  H  or  V  but  is  parallel  to  H  or  V.     The  method  by 


REVOLUTION  OF  POINTS  41 

which  this  may  be  done  may  be  readily  imderstood  by  con- 
sidering that  a  secondary  H  ohV  plane  passes  through  the  axis. 
The  revolution,  then,  may  be  referred  to  this  new  H  or  V  plane 
and  may  be  accompHshed  as  indicated  in  Article  32. 

Fig.  39  shows  the  plan  and  elevation  of  a  line  MN  which  is 
to  be  revolved  into  the  plane  of  the  axis  PQ.  The  revolution 
may  be  accomplished  as  indicated  above  or  by  working  the 


Fig.  40. 

problem  without  the  use  of  the  secondary  H  or  V  plane.  In  the 
figure  the  line  MN  has  been  revolved  into  the  plane  PQ  directly, 
M  falls  at  a  point  whose  elevation  is  m'',  m''  being  a  distance 
from  the  axis  equal  to  the  hypotenuse  of  a  triangle  whose  base  is 
the  distance  of  m'  to  the  axis,  or  m'y',  and  whose  altitude  is  the 
distance  M  is  from  the  plane  of  the  axis,  or  mx. 

36.  To  revolve  a  plane  into  H  or  V  about  the  H  or  V  trace. 
Assume  a  point  on  the  trace,  as  A  on  the  trace  t'T  in  Fig.  40. 


42 


ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 


Revolve  this  point  into  H  about  tT  as  an  axis  as  in  Article  32. 
Through  this  revolved  position,  Ah,  and  the  point  T  on  the  G.  L. 
draw  the  line  t'nT.  This  will  be  the  revolved  position  of  the  V 
trace  as  it  lies  on  H  and  t'nTt  is  the  revolved  position  of  the 

plane  T. 

To  revolve  the  plane  into  V  a  point  on  the  H  trace  should  be 
taken  and  the  V  trace  used  as  an  axis. 


Fig.  41. 

37.  To  find  the  plan  and  elevation  of  a  line  in  a  plane  when  the 
revolved  position  of  the  line  and  plane  are  given. 

In  Fig.  41  is  shown  a  plane  T  and  a  Une  MN  in  the  plane  T  as 
they  lie  revolved  into  the  H  plane  at  tTt'n  and  MrNh-  Xo  find 
the  actual  position  of  the  traces  and  the  plan  and  elevation  of 
the  line  assume  Ph,  the  revolved  position  of  a  point  on  the  re- 
volved V  trace.  P,  since  it  is  on  the  V  trace,  will  lie  in  V  and  its 
plan  view  therefore  lies  on  the  G.  L.    Through  Ph  draw  a  per- 


REVOLUTION  OF  POINTS  43 

pendicular  to  Tt,  since  Tt  must  be  the  axis  of  revolution,  and 
extend  it  to  the  G.  L.  at  p.  p,  then,  is  the  plan  of  a  point  P  on 
the  V  trace,  and  the  elevation  will  be  at  p'  which  is  below- the 
G.  L.  a  distance  equal  to  ox.  To  find  ox  lay  off  px  equal  to  opn- 
ox  then  is  the  base  of  the  triangle  whose  hypotenuse  is  px,  or 
oPh.  Draw  t'T  through  T  and  p'  and  it  will  be  the  true  position 
of  the  V  trace  and  tTt'  will  represent  the  true  position  of  the 
plane. 

To  find  the  plan  and  elevation  of  any  line  MN  in  this  plane 
whose  revolved  position  is  MrNh  extend  MrNh  to  Ah  on  the 
H  trace  and  Bh  on  the  revolved  V  trace.  When  the  plane  is 
revolved  back  to  its  true  position  tTt',  Bh  will  fall  at  b'  so  that 
Tb'  is  equal  to  BhT  and  Ah  falls  at  Ah-  A  is  shown  in  elevation 
at  a'  on  the  G.  L.  and  B  is  shown  in  plan  at  b,  also  on  the  G.  L. 
Anb  is  the  plan  and  a'b'  is  the  elevation  of  the  required  line 
extended,  and  M  and  N  may  be  located  on  this  line. 

PROBLEMS  ON  REVOLUTION 

85.  M  =  o;  4-1;  o.  N  =  +3;  +4;  o.  O  =  +1;  +2;  +2.  Re- 
volve O  into  H  using  MN  as  an  axis.  Find  the  length  of  the  radius  of 
revolution. 

86.  A  =  o;  o; +1.  B  =  +3;  o; +4-  0  =  +2; -i; -i.  Using  AB 
as  an  axis  revolve  O  into  V.     Give  the  length  of  the  radius  of  revolution. 

87.  M  =  o;  —2;  —2.  N  =  4-3;  — i;  —I.  Revolve  MN  into  H  about 
its  own  plan  view. 

88.  A  =  — 3;o;o.  B  =  +1;  —2;  —2.  Revolve  AB  into  V  about 
a'b'  as  an  axis. 

89.  A  =  +1;  —  2;  — o.  B  =  —2;  —  i;  —3.  Revolve  AB  into  H  about 
ab. 

90.  The  plane  S  =  —  2;  +30;  —45.    Revolve  plane  S  into  H  about  sS. 

91.  The  plane  T  =  -|-i;  -I-120;  —135.  Revolve  plane  T  into  V  about 
its  V  trace. 

92.  The  plane  S  =  o;  +45;  —60.  Revolve  this  plane  into  H  and  in 
this  position  locate  a  i"  square  hole  in  this  plane.  Show  the  plan,  elevation, 
and  end  view  of  the  hole  after  the  plane  is  returned  to  its  original  position. 

93.  The  point  T  =  o;  o;  o.  The  point  t'  =  +3;  o;  —3.  The  point 
tv  =  o;  o;  —3.  t'Ttv  is  the  revolved  position  of  a  plane  T  as  it  Ues  in  V. 
Show  its  true  position. 

94.  The  point  S  =  o;  o;  o.  The  point  s  =  +2;  —3;  o.  The  point 
s'h  =  —  I ;  —  3;  o.  sSs'h  is  the  position  of  a  plane  S  as  it  Hes  revolved  into 
H.    Show  it  in  its  true  position. 


44  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

95^  The  plane  T  =  qo  ;  —2;  —3.  The  point  0  which  lies  in  T  lies  at 
the  point  o;  x,  —  i.  Find  the  plan  and  elevation  of  the  point  and  revolve 
into  V  about  t'T. 

96.  Plane  S  =  o;  —60;  +30.  The  point  0  lies  in  this  plane  at  —2; 
—  I ;  X.  Show  a  plan  and  elevation  of  this  point  and  revolve  it  into  H  about 
tT. 


CHAPTER   VII 
PROBLEMS    ON   THE   LINE 

38.  Proposition  i.     Given  two  views  of  a  line  to  find  its  length. 

Discussion.  Revolve  the  Une  parallel  to  any  of  the  planes  of 
projection.  In  this  position  the  Une  will  be  projected  on  the 
plane  to  which  it  is  parallel  in  its  true  length.  Therefore,  the 
length  of  the  projection  is  the  length  of  the  line  itself. 

Construction.  First  Method.  Let  it  be  required  to  find  the 
over-all  length  of  the  brace  AB  shown  in  the  drawing  of  the 
bracket  in  Fig.  42.  Using  the  center  Une  of  the  vertical  post  as 
an  axis  turn  the  brace  until  it  becomes  parallel  to  the  V  plane 
and  takes  the  position  shown  by  the  light  fines.  Both  A  and  B 
will  describe  arcs  of  circles  during  the  revolution,  and  since  the 
planes  of  these  arcs  are  perpendicular  to  the  axis  of  the  revolu- 
tion (Art.  32),  the  points  will  remain  the  same  distance  below 
H.  In  the  new  position  shown  by  a"  and  b''  the  fine  is  paraUel 
to  V  and  the  distance  a''b"  is  the  true  length  of  the  brace. 

Construction.  Second  Method.  In  Fig.  43  the  same  problem 
is  solved  by  revolving  AB  parallel  to  H  by  using  ab  as  an  axis 
of  revolution.  The  revolved  position  of  A  is  found  at  Ah  by 
Article  32.  Ana  is  equal  to  the  distance  from  a'  to  the  G.  L.  and 
is  drawn  perpendicular  to  ab.  In  fike  manner  B  is  found  at  Bh- 
The  fine  ArBh  being  paraUel  to  H  is,  of  course,  equal  in  length 
to  AB. 

Corollary.  Given  the  length  of  a  line  and  one  view  of  it  to 
construct  the  other  view,  when  the  position  of  some  point  on  the 
fine  is  known. 

Construction.  Let  the  problem  be  to  make  a  drawing  of  the 
hopper  (Fig.  44)  whose  plan  is  abed  —  mnop.  The  length  of 
the  edge  BN  is  12",  and  the  plane  of  the  opening  ABCD  fies 
4"  below  H. 

45 


46 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


First  draw  the  line  a'b'c'  4"  below  the  G.  L. ;  this  is  the  eleva- 
tion of  the  opening  ABCD.  Now  with  b'  as  a  center  strike  an 
arc  whose  radius  is  12"  \  somewhere  in  this  arc  will  lie  n'',  the 


Fig.  42. 


elevation  of  the  revolved  position  of  N.  To  find  this  point  re- 
volve BN  parallel  to  G.  L.  at  bn,.  From  n,  drop  a  perpendicu- 
lar to  intersect  the  arc;  the  point  thus  found  is  n",  the  elevation 


PROBLEMS  ON  THE  LINE 


47 


\ 

o'\     \ 

WWvwJ 

Fig.  43- 


48 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


of  the  revolved  position  of  N.  BN  may  now  be  revolved  to  the 
first  position  bn^  and  n"  will  be  shown  at  n'.  The  elevation  of 
the  hopper  may  now  be  drawn  as  shown  by  the  figure. 

39.   Special  Cases,     i.   To  find  the  length  of  a  Hne  which  lies 
in  the  first  and  third  quadrants. 


2.   To  find  the  length  of  a  hne  which  lies  in  the  profile  plane. 

40.  Proposition  2.  Given  two  views  of  a  Hne  to  find  where  it 
pierces  the  planes  of  projection. 

Discussion.  The  point  where  the  Hne  pierces  H  must  lie  on 
the  line  itself  and  in  the  H  plane.  Therefore,  the  projections  of 
the  required  piercing  point  will  he  on  the  projections  of  the  given 


PROBLEMS   ON   THE   LINE 


49 


line,  and  the  elevation  will  lie  on  the  G.  L.  If  the  elevation  of 
the  line  be  extended  to  meet  the  G.  L.  this  point  will  be  the  ele- 
vation of  the  required  piercing  point;  its  plan  view  will  lie  on  the 


plan  view  of  the  line  and  in  a  perpendicular  from  the  elevation. 
In  a  similar  manner  the  V  piercing  point  may  be  found. 

Construction.  In  Fig.  45  let  the  problem  be  to  find  the  plan 
and  elevation  of  the  point  where  the  center  line  OP  of  the  timber 
pierces  H  and  V.     By  extending  o'p'  to  the  G.  L.  the  point  m' 


50  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

is  found;  m'  is  the  elevation  of  the  point  where  the  line  OP 
pierces  the  H  plane.  The  plan  view  of  this  H  piercing  point  will 
be  on  op  where  the  perpendicular  from  m'  intersects  op  at  m; 
since  M  is  in  H  (as  shown  by  the  elevation  m')  and  on  OP  (as  m 
and  m'  show)  it  will  be  the  point  where  OP  pierces  H. 

To  find  where  OP  pierces  V,  op  is  extended  to  the  G.  L.  at  n, 
and  n'  is  located  on  o'p'  as  shown.  The  point  N,  being  in  V  and 
on  the  line  OP,  will  be  the  point  where  OP  pierces  V  and  n  is  the 
plan  and  n'  the  elevation  of  this  point. 

In  a  similar  manner  the  P  piercing  point,  if  it  be. required,  may 
be  found. 

41.  Special  Cases,  i.  Given  the  location  of  the  H  and  V 
piercing  points  of  a  line  and  the  length  of  the  line  to  draw  its 
plan  and  elevation. 

2.  Find  the  piercing  points  of  a  line  which  lies  in  a  profile 
plane. 

PROBLEMS  ON  THE  LINE 

97.  Draw  the  plan  and  elevation  of  a  triangular  pyramid  whose  base  is 
A  =  o;  — i;  — i;  B  =  +2;  —3;  — i;  C  =  +ii;  —2;  — i,  and  whose 
apex  isO=  +1;  —2;  —4.    Find  the  true  length  of  OA. 

98.  The  center  line  of  a  pipe  intersects  the  floor  of  a  room  at  the  point 
A  =  -36";  +48";  +0",  and  intersects  the  front  wall  at  B  =  -36";  o"; 
+36''.    Find  how  long  the  pipe  is  between  these  points. 

99.  The  center  line  of  a  timber  is  so  located  that  one  end  is  at  the  point 
M  =  o;  +36";  +6",  and  the  other  end  is  at  the  point  N  =  —48";  —36"; 
—36".    Find  how  long  the  timber  is. 

100.  A  brace  rests  against  a  floor  at  o;  —24";  o,  and  against  the  wall  at 
—36";  o;  —48".  Find  the  true  length  of  the  brace  (center  line  only  con- 
sidered) between  these  points. 

loi.  The  center  line  of  a  tunnel  is  500'  long.  One  portal  lies  at  the  point 
o;  —  o';  —40',  and  from  this  point  the  tunnel  runs  north  60  degrees  east  on 
a  rising  6  per  cent  grade.  Show  its  plan  and  elevation  and  find  the  actual 
distance  between  portals. 

Note.  In  all  land  measurement  distances  are  measured  on  the  horizon- 
tal. Thus,  the  length  of  the  tunnel  given  as  500'  is  really  the  length  of  its 
H  projection  or  plan  view,  and  the  distance  asked  for  is  the  actual  linear 
distance  from  one  portal  to  the  other. 

102.  A  drain  starts  at  a  point  —48";  —36";  o,  and  runs  north  45  degrees 
east  for  96"  on  a  falling  10  per  cent  grade.  How  many  inches  of  drain  pipe 
will  be  required? 


PROBLEMS  ON  THE  LINE  5 1 

103.  A  prop  10'  long  rests  on  a  floor  6'  in  front  of  a  wall.  Find  where  it 
will  rest  against  the  wall  when  its  plan  view  incHnes  60  degrees  to  the  floor 
line. 

104.  A  pipe  hes  in  a  side  wall  so  that  it  slopes  30  degrees  to  the  horizontal 
and  goes  through  the  ceiUng  6'  back  of  the  front  wall.  Find  where  it  cuts 
the  front  wall  when  the  pipe  is  8'  long. 

105.  A  tunnel  starts  from  the  bottom  of  a  shaft  at  o;  —100';  —65',  and 
runs  south  60  degrees  east  on  a  rising  15  per  cent  grade.  Find  the  point 
where  the  tunnel  will  reach  the  surface  and  how  long  it  will  actually  be. 

106.  A  brace  rests  against  a  wall  60"  above  the  ground,  and  rests  against 
the  ground  60''  in  front  of  the  wall.  The  brace  is  108"  long.  Show  three 
views  of  it. 

107.  How  many  feet  of  pipe  will  be  required  to  join  an  opening  in  a  floor 
18"  back  of  the  front  wall  of  a  building  with  an  outlet  in  the  left  side  wall 
60"  below  the  floor  and  36"  back  of  the  front  wall?  The  opening  in  the 
floor  is  48"  from  the  side  wall. 


CHAPTER   VIII 
PROBLEMS    ON   THE   PLANE 

42.  Proposition  3.  To  find  the  traces  of  a  plane  when  two 
views  of  any  of  its  lines  are  given. 

Discussion.  Since  the  trace  of  a  plane  is  the  line  in  which  the 
plane  cuts  H  or  V  it  must  contain  the  points  in  which  all  lines 
contained  in  the  plane  cut  H  or  V.  Therefore,  if  the  H  and  V 
piercing  points  of  the  given  lines  be  found  these  will  determine 
the  traces  of  the  plane  of  the  Unes. 

Construction.  Let  it  be  required  in  Fig.  46  to  find  the  traces 
of  the  plane  of  that  side  of  the  hood  given  by  the  two  hnes  AB 
and  BC.  By  Proposition  2  find  where  AB  and  BC  pierce  H  and 
V;  AB  piercing  H  at  M  and  V  at  N;  BC  piercing  H  at  O  and  V 
at  P.  Join  M  and  0;  this  will  give  the  H  trace  tT.  Join  N  and 
P;  this  gives  the  V  trace  t'T.  The  plane  of  the  two  lines  then 
is  tTt'.     The  traces  must  of  course  meet  on  the  G.  L. 

43.  Special  Cases.    Find  the  traces  of  the  plane  of  two  lines: 
J.   When  one  line  is  parallel  to  H. 

2.  When  one  line  is  parallel  to  H  and  the  other  parallel  to  V. 

3.  When  one  line  is  parallel  to  the  G.  L. 

4.  When  one  line  is  parallel  to  the  G.  L.  and  the  other  is 

parallel  to  H  or  V. 

5.  When  the  two  lines  are  given  by  their  plans  and  end  views. 
Note.     In  working  the  above  special  cases  it  should  be 

remembered  that  when  a  line  lies  in  a  plane  and  is  parallel  to 
H  or  V,  it  will  be  parallel  to  the  H  or  V  trace  of  the  plane.  See 
observations  on  planes,  Article  23. 

44.  Proposition  4.  Given  the  plan  and  elevation  of  a  point 
to  pass  through  it  a  plane  parallel  to  two  lines  whose  plans  and 
elevations  are  given. 

52 


PROBLEMS   ON  THE   PLANE 


S3 


Discussion.  If  through  the  given  point  two  lines  be  drawn 
parallel  to  the  two  given  lines,  the  plane  of  these  two  lines  will 
be  the  required  plane. 


Construction.  Let  the  problem  be  to  cut  off  the  trough, 
Fig.  47,  one  of  whose  edges  is  NP,  by  a  plane  through  P  parallel 
to  MN  and  NO.  Through  p  draw  pq  and  pr  parallel  to  the  plan 
of  the  two  given  lines,  mn  and  no.  Also  draw  p'q'  and  pV 
parallel  to  m'n'  and  n'o'.  Find  now  where  these  Unes,  PQ  and 
PR,  which  are  parallel  to  the  given  lines,  pierce  H  and  V.  PQ 
pierces  H  at  X  and  V  at  q'.  PR  pierces  V  and  y'.  Draw  q'y' 
and  extend  it  to  T;    also  draw  Tx.     These  are  the  H  and  V 


54 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


traces  respectively,  and  since  PR  is  parallel  to  H  the  trace  tT 
should  be  parallel  to  pr.  The  plane  T  then  is  the  required  plane 
through  P  parallel  to  MN  and  MO. 


45.  Special  Cases.    Pass  a  plane  through  a  point  parallel  to 
two  given  lines: 

1.  When  one  of  the  lines  is  parallel  to  the  G.  L. 

2.  When  one  of  the  lines  is  parallel  to  the  H  or  V. 

3.  When  one  line  is  parallel  to  H  and  the  other  is  parallel  to  V. 

4.  Pass  a  plane  through  one  Hne  parallel  to  another. 

46.  Proposition  5.     Given  the  plan  and  elevation  of  a  point 
to  pass  through  it  a  plane  parallel  to  a  given  plane. 

Discussion.     Since  the  required  plane  is  to  be  parallel  to  the 
given  plane  the  corresponding  traces  of  these  two  planes  will  be 


PROBLEMS  ON  THE  PLANE 


55 


parallel.  The  direction  of  the  required  traces  is,  therefore, 
known.  To  find  their  location  a  line  may  be  drawn  through  the 
given  point  parallel  to  either  trace  of  the  given  plane  and  its  H 


Fig.  48. 

and  V  piercing  point  found.     With  one  point  on  either  trace  and 
the  direction  of  the  traces  known  the  plane  may  be  located. 

Construction.     In  Fig.  48  is  shown  the  plan  and  elevation  of 
an  octagonal  prism  whose  base  lies  in  the  plane  T.      The  prob- 


56 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


lem  is  to  cut  off  the  prism  by  a  plane  parallel  to  T  through  the 

point  O. 

Through  0  draw  a  line  OP  parallel  to  t'T.  This  line  pierces 
H  at  P.  Through  P  draw  sS  parallel  to  tT  and  through  S  draw 
s'S  parallel  to  t'T;  sSs'  is  the  required  plane  through  O  parallel 
to  the  plane  T. 


Fig.  49. 

47.  Special  Case.  Pass  a  plane  through  a  point  parallel  to  a 
plane  which  is  parallel  to  the  G.  L. 

48.  Proposition  6.  Given  the  plan  and  elevation  of  a  line  to 
draw  through  a  point,  whose  plan  and  elevation  are  given,  a 
plane  perpendicular  to  the  line. 

Discussion.  When  a  line  is  perpendicular  to  a  plane  the  plan 
and  elevation  of  the  line  are  perpendicular  to  the  corresponding 
traces  of  the  plane.  Thus  the  direction  of  the  traces  of  the 
required  plane  is  known.     If  then  a  line  be  drawn  through  the 


PROBLEMS  ON  THE  PLANE  57 

given  point  parallel  to  the  direction  of  either  of  these  traces  this 
line  will  be  in  the  required  plane.  By  finding  its  H  or  V  piercing 
point  the  location  of  one  point  on  one  of  the  traces  may  be  found 
and  the  plane  located. 

Construction.  Let  the  problem  be  to  find  the  traces  of  a  plane 
through  the  point  0  which  will  cut  a  section  perpendicular  to 
MN,  the  axis  of  the  pyramid  in  Fig.  49.  The  H  trace  of  the 
required  plane  will  be  perpendicular  to  the  plan  view  of  the  axis, 
and  the  V  trace  will  be  perpendicular  to  the  elevation.  There- 
fore, through  O,  the  given  point,  draw  a  line  parallel  to  the 
direction  of  the  required  H  trace;  its  plan  view  will  be  op  and 
its  elevation  o'p'.  This  Hne  OP  is  a  line  of  the  required  plane 
since  it  contains  a  point  in  the  required  plane  and  is  parallel  to 
one  of  the  traces  of  that  plane.  It  pierces  V  at  p'  and  if  through 
p'  r'R  be  drawn  perpendicular  to  m'n'  the  elevation  of  the  axis, 
it  will  be  the  V  trace  of  the  required  plane.  The  H  trace  is  Rr 
which  is  perpendicular  to  the  plan  view  of  the  axis.  rRr',  then, 
is  the  required  plane. 

49.  Special  Cases,  i .  Construct  this  same  problem  when  the 
point  does  not  He  on  the  line  to  which  the  plane  is  to  be  per- 
pendicular. 

2.  Find  the  location  of  a  plane  which  is  parallel  to  a  given 
plane  at  a  given  distance  from  it. 

PROBLEMS  ON  THE  PLANE 

108.  The  point  A  =  o;  —  2;  — 3;  the  point  B  =  +1;  — i;  — i;  the 
point  C  =  -f-2;  —3;  —2.    Find  the  traces  of  the  plane  of  the  triangle  ABC. 

109.  The  point  A  =  -f  i;  — i;  —3;  the  point  B  =  -f  i;  —2;  — i;  the 
point  M  =  +3;  — i;  —3;  the  point  N  =  -I-3;  —2;  — i.  Pass  a  plane 
through  AB  and  MN  and  show  its  traces. 

no.  The  line  AB  is  parallel  to  the  G.  L.  1"  below  H  and  2"  back  of  V. 
The  hne  MN  is  parallel  to  the  G.  L.  i"  above  H  and  2"  in  front  of  V.  Find 
the  traces  of  the  plane  of  these  two  lines. 

111.  The  point  M  =  — i;  +2;  +2;  the  point  N  =  —2;  —  i;  — i;  the 
point  O  =  —2;  +2;  — i;  the  point  P  =  o;  — i;  —2.  Pass  a  plane  through 
P  parallel  to  the  lines  MN  and  ON. 

112.  The  point  A  =  o;  —  i;  o;  the  point  B  =  +2;  —  i;  —3;  the  point 
C  =  +2;  —3;  —3;  the  point  O  =  —2;  —3;  — i.  Pass  a  plane  through 
O  parallel  to  AB  and  BC. 


58  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

113.  The  plane  T  =  00  ;  —  2;  —4;  the  point  0  =  —3;  —2;  —3.    Pass 
a  plane  through  0  parallel  to  plane  T. 

114.  The  plane  S  =  o;  +60;  —60;  the  pomt  0  =  +2;  —  i;  —3.    Pass 
a  plane  through  0  parallel  to  plane  S. 

11$.  The    point    M  =  — i;  —2;  — i;     the    pomt    N  =  —2;  —3;  —2. 
Pass  a  plane  through  M  perpendicular  to  the  line  MN. 

116.  The  point  A  = —i; —3;  —  4;  the  point  B  =  — 3;  — i;  — i.    Pass 
plane  through  AB  perpendicular  to  it  at  its  middle  point. 

117.  The  point  M  =  o;  o;  o;   the  p^oint  N  =  —  i;  —3;  —3;    the  point 
0=— 3;— i;— 2.    Pass  a  plane  through  0  perpendicular  to  MN. 


CHAPTER   IX 
PROBLEMS   ON  ANGLES 

50.  Proposition  7.  Given  two  views  of  an  angle  to  find  its 
true  size. 

Discussion.  If  the  plane  of  the  angle  be  revolved  into  co- 
incidence with  H  or  V  the  angle  in  this  position  will  be  shown  in 
its  true  size.  Therefore,  find  the  plane  of  the  angle,  and  using 
either  the  H  or  the  V  trace  as  an  axis  revolve  the  angle  into  H  or 
V.  When  the  angle  coincides  with  H  or  V  it  will  be  shown  in 
its  true  size. 

Construction.  In  Fig.  50  let  the  problem  be  to  find  the  true 
size  of  all  the  angles  between  the  edges  of  the  section  of  the  tri- 
angular prism;  or,  in  other  words,  the  shape  and  size  of  the 
triangle  ABC.  By  Proposition  3  find  the  plane  of  the  triangle; 
its  traces  are  rR  and  r'R.  With  the  V  trace  r'R,  as  an  axis, 
revolve  the  triangle  into  V  by  the  method  in  Article  36.  When 
coinciding  with  V  the  triangle  will  occupy  the  position  AyByCv, 
and  as  this  is  its  true  shape  and  size  the  edges  and  the  angles  may 
be  measured  with  scale  and  protractor. 

51.  Corollary.  Given  two  views  of  one  side  of  an  angle,  the 
traces  of  its  plane,  and  its  size,  to  construct  the  views  of  the 
other  side. 

Discussion.  If  the  given  side  be  revolved  into  H  or  V  about 
the  trace  of  the  given  plane,  the  angle  may  be  constructed  in  this 
position  in  its  true  size.  If  now  the  plane  be  revolved  back  to 
its  original  position  the  two  views  of  the  required  angle  may  be 
found. 

Construction.  Let  the  problem  be,  in  Fig.  51,  to  construct 
a  rectangular  opening  in  the  plane  R  with  MN  as  one  edge  and 
the  other  edge  one-half  as  long.  By  Article  34  revolve  MN  about 
rR  into  H.  In  this  position  MN  will  lie  at  MrNh  and  the  plane 
R  will  lie  at  rRr'n-     With  MhNh  as  one  side  construct  the  rec- 

59 


6o 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


tangle  MhNhOhPh  making  MrPh  equal  to  i  MrNh.  This 
rectangle  is  the  true  shape  and  size  of  the  required  rectangular 
opening. 

Now  revolve  rRr'n  back  to  its  first  position  rRr'  and  with  it 
revolve  the  rectangle  MhNhOhPh-    To  find  the  plan  of  P  draw 


Fig.  so. 

the  diagonal  NrPh  and  let  it  intersect  the  trace  rR  at  a.  Now 
when  the  plane  R  returns  to  its  first  position  Nh  revolves  to  n 
and  the  diagonal  aBn  revolves  to  ab.  Since  Ph  is  on  this 
diagonal  p,  the  required  plan  view  of  P  will  be  found  at  the 
point  where  a  perpendicular  from  Ph  to  rR  intersects  ab.  To 
find  the  elevation  of  P  draw  first  a'b',  a'  being  on  the  G.  L.  since 


PROBLEMS   ON  ANGLES 


6i 


A  is  on  the  trace  rR,  and  b'  being  on  r'R.  p',  the  elevation  of 
P,  will  be  found  on  a'b'  in  a  perpendicular  to  the  G.  L.  from  p. 
In  like  fashion,  O  may  be  found  and  the  plan  and  elevation  of  the 
required  rectangular  opening  constructed. 


Fig.  si. 


52.  Corollary.  Given  a  line  and  a  point  to  draw  through  the 
point  a  Une  making  a  given  angle  with  the  given  Une. 

Discussion.  If  a  plane  be  passed  through  the  given  point  and 
line  it  will  be  the  plane  of  the  required  angle.  If  this  plane  be 
revolved  into  H  or  V  the  true  relation  of  the  point  to  the  line 
will  be  shown  when  they  are  in  the  new  position.  The  given 
angle  may  then  be  constructed.     If  now  the  plane  be  revolved 


62  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

back  to  its  original  position  the  plan  and  elevation  of  the  required 
angle  may  be  found. 

Construction.  Let  the  construction  be  made  according  to 
the  discussion. 

53.  Special  Cases,  i.  To  find  the  size  of  an  angle  when  one 
side  is  parallel  to  the  G.  L. 

2.   To  draw  two  views  of  the  bisector  of  an  angle. 

PROBLEMS   ON  ANGLES 

118.  Given  the  triangle  ABC,  A  =  o;  —3;  —2;  B  =  — i;  — i;  — i; 
C  =  —2;  —2;  —3,  to  find  its  true  shape  and  size. 

119.  The  point  A  =  o;  o;  —3;  the  point  B  =  +1;  o;  —  i;  the  point 
C  =  +2;  o;  —2;  the  point  O  =  —  i;  —3;  —  i.  Find  the  true  shape  and 
size  of  each  of  the  faces  of  the  triangular  pyramid  whose  base  is  ABC  and 
upon  whose  apex  is  0. 

120.  The  point  M  =  o;  —3;  — i;  the  point  N  =  o;  — r;  —3;  the  point 
0=  —2;  — i;  —I.    Find  the  true  shape  and  size  of  the  angle  MNO. 

121.  The  plane  S  =  o;  —60;  +45-  The  point  O  in  this  plane  lies  at 
4-2;  +1;  +x.  With  O  as  one  corner  construct  an  equilateral  triangular 
hole  in  the  plane  S  with  sides  1"  long.  Let  the  base  of  the  triangular  hole 
be  parallel  to  the  H  trace  of  the  plane. 

122.  The  plane  T=— 3;— 45;+6o.  A  line  lies  in  this  plane  at  M  = 
—  5;  — x;  —I  and  N  =  —6;  —2;  y.  With  MN  as  one  side  construct  a 
square  hole  in  plane  T. 

123.  The  plane  S  =  o;  +45;  —60.  The  point  O  =  +2;  —  i;  x. 
Through  the  point  O,  which  Hes  in  plane  S,  draw  a  line  in  plane  S  perpen- 
dicular to  the  H  trace  of  S  and  find  the  angle  it  makes  with  the  V  plane. 

124.  M  =  —3;  —2;  —2.  N  =  o;  — i;  — i.  O  =  —2;  — i;  —3. 
Draw  a  Une  through  0  making  an  angle  of  30  degrees  with  MN.  (Two 
solutions  possible.) 

125.  M  =  -3; -i; -I.  N  =  o;-2;-3.  0  =  -2;  -4; -i. 
Through  0  draw  a  line  perpendicular  to  MN. 

126.  A  =  -i;  +2;  +2.  B  =  -i;  +3;  +4.  O  =  o;  +1;  +1. 
Through  O  draw  a  line  perpendicular  to  AB. 

127.  A  =  o; +1; +4.  B  =  o;— i;— 2.  C  =  +2;  — 3;  — i.  Show 
three  views  of  the  bisector  of  the  angle  ABC. 

128.  An  observer  stands  in  a  lighthouse  tower  at  A  =  +1000';  +100'; 
—300',  and  at  10  o'clock  observes  a  ship  N.  30°  E.  from  him  at  an  angle  of 
depression  of  30  degrees.  One  hour  later  the  same  ship  is  N.  60°  W.  from 
him  at  an  angle  of  depression  of  15  degrees.  Assuming  the  ship  keeps  the 
same  course  and  speed  how  will  it  bear  from  him  and  what  will  be  its  angle 
of  depression  at  1 2  o'clock? 


CHAPTER  X 
PROBLEMS    ON   POINTS,   LINES,   AND   PLANES 

54.  Proposition  8.  To  draw  the  plan  and  elevation  of  the 
intersection  of  two  given  planes. 

Discussion.  Since  the  line  of  intersection  contains  all  of  the 
points  common  to  both  planes,  its  location  will  be  determined 
if  any  two  of  these  common  points  be  located.     If,  then,  a  line 


be  drawn  from  the  intersection  of  the  H  traces  to  the  intersection 
of  the  V  traces,  the  plan  and  elevation  of  this  line  show  the 
required  intersection. 

Construction.  Let  T  and  S,  Fig.  52,  be  the  given  planes 
whose  intersection  is  required.  The  H  traces  intersect  at  a  point 
whose  plan  is  a  and  whose  elevation  is  a'.  The  V  traces  intersect 
at  a  point  whose  plan  is  b  and  whose  elevation  is  b'.  ab,  then, 
is  the  plan  and  a'b'  is  the  elevation  of  the  required  intersection 
AB. 

63 


64 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


In  case  the  traces  of  the  given  planes  do  not  intersect  con- 
veniently on  the  drawing,  Fig.  53  shows  a  method  of  finding 
the  required  intersection  without  using  the  intersection  of  the 
traces.  T  and  S  are  again  the  given  planes,  Rr  is  an  auxiliary 
plane  parallel  to  V,  which  cuts  the  Une  AB  from  the  plane  S,  and 
the  line  BC  from  the  plane  T,  thus  giving  a  point  B  common  to 
both  planes.    In  like  manner  a  second  point  N  on  the  required 


intersection  may  be  found  by  passing  an  auxiUary  plane  U 
through  S  and  T  parallel  to  H.  B  and  N  when  joined  give  the 
plan  view  bn  and  the  elevation  b'n'  of  the  required  intersection 
of  the  planes  S  and  T. 

55-  Special  Cases,  i.  Find  the  intersection  of  two  planes 
when  one  pair  of  traces  are  parallel. 

2.  Find  the  intersection  of  two  planes  which  are  parallel  to 
the  G.  L. 

56.  Proposition  9.  Given  a  plan  and  elevation  of  a  line  to 
find  where  it  pierces  a  given  plane. 


PROBLEMS  ON  POINTS,  LINES,  AND  PLANES 


6S 


Discussion.  If  any  plane  be  passed  through  the  given  line  it 
will  contain  all  points  of  that  line.  It  will,  therefore,  contain 
the  required  point.  Since  the  required  point  lies  in  the  given 
plane  and  in  the  auxiliary  plane  it  will  lie  on  their  line  of  inter- 
section.   If  then  the  line  of  intersection  be  found  and  the  point 


\           m^X 

\ 

^/           \ 

\. 

\  A 

n 

\A 

-/' 

Fig.  54. 

common  to  it  and  the  given  line  be  located  it  will  be  the  required 
piercing  point. 

Construction.  In  Fig.  54  is  shown  the  plan  and  elevation  of 
a  timber  whose  center  line  is  OP.  The  problem  is  to  find  the 
point  where  this  line  OP  pierces  the  plane  tTt'.     Pass  the  auxil- 


66  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

iary  plane  R  through  OP.  Since  the  required  piercing  point  lies 
in  the  plane  R  and  in  the  given  plane  T  it  will  He  on  their  inter- 
section MN.  (See  Proposition  8.)  The  required  point  lies  on 
MN  and  OP;  therefore,  it  will  be  at  the  intersection  of  these  two 
lines  or  at  Q.  q  then  is  the  plan  and  q'  is  the  elevation  of  the 
required  point. 

It  should  be  noted  that  any  auxiliary  plane  through  OP  will 
serve,  but,  for  convenience  in  drawing,  one  which  is  perpendicular 
to  H  or  V  should  be  used.  The  points  where  the  edges  of  the 
timber  pierce  T  may  be  found  in  a  similar  manner  and  when 
joined  give  the  plan  and  elevation  of  the  section  cut  from  the 
timber  by  plane  T. 

57.  Special  Cases.     Find  where  a  line  pierces  a  plane: 

1.  When  the  line  is  parallel  to  the  G.  L. 

2.  When  the  plane  is  parallel  to  the  G.  L. 

3.  When  the  plane  is  parallel  to  the  G.  L.  and  the  Hne  lies  in 

a  profile  plane. 

4.  When  the  plan  and  elevation  of  the  line  are  parallel  respec- 

tively to  the  H  and  V  traces  of  the  given  plane. 

58.  Proposition  10.  Given  the  plan  and  elevation  of  a  point 
to  find  its  shortest  distance  from  a  given  plane. 

Discussion.  The  shortest  distance  from  a  point  to  a  plane  is 
the  perpendicular  distance  from  the  point  to  the  plane.  If  the 
perpendicular  be  drawn  from  the  point  to  the  given  plane  and 
the  point  where  the  perpendicular  pierces  the  plane  be  found, 
the  distance  between  these  two  points  will  be  the  required 
shortest  distance. 

Construction.  In  Fig.  55  the  problem  is  to  find  the  altitude 
of  the  pyramid  whose  base  lies  in  the  plane  T  and  whose  apex  is 
at  0.  From  O  draw  a  line  perpendicular  to  the  plane  T.  (See 
Observation/,  Article  23.)  Find  where  this  line  pierces  plane  T 
by  Article  56.  p  is  the  plan  and  p'  the  elevation  of  this  piercing 
point.  Then  op  is  the  plan  and  o'p'  is  the  elevation  of  the 
shortest  distance  from  the  apex  of  the  pyramid  to  the  plane  of 
its  base.  '  Its  true  length  may  be  found  by  Article  38,  as  shown 
at  o'^p'. 

It  should  be  noted  that  P,  which  is  the  foot  of  the  perpendicu- 


PROBLEMS  ON  POINTS,  LINES,  AND   PLANES 


67 


lar  drawn  from  O  to  the  plane  T,  is  the  projection  of  the  point 
O  on  the  oblique  plane  T. 


Fig.  55. 

59.  Corollary.  Given  a  plane,  the  plan  and  elevation  of  the 
projection  of  a  point  upon  this  plane,  and  its  distance  from  the 
plane  to  find  the  plan  and  elevation  of  the  point. 

Discussion.  If  a  perpendicular  be  erected  to  the  given  plane 
through  this  given  projection  of  the  point,  the  required  point 


68  ESSENTIALS   OF   DESCRIPTIVE   GEOMETRY 

will  lie  in  this  line.  To  find  its  location  a  point  on  the  perpen- 
dicular must  be  found  at  the  given  distance  from  the  plane. 

Construction.  Let  the  construction  be  made  in  accordance 
with  the  suggestions  in  the  discussion. 

60.  Special  Cases,  i .  Find  the  shortest  distance  from  a  point 
to  a  plane  when  the  plane  is  parallel  to  the  G.  L. 

2.  Given  two  parallel  planes  to  find  the  shortest  distance 
between  them. 

PROBLEMS 

129.  The  plane  S  =  o;  —45;  +60.  The  plane  T  =  +3;  —300;  +300- 
Find  the  true  length  of  that  portion  of  the  intersection  of  planes  S  and  T 
which  lies  between  H  and  V. 

130.  The  plane  S  =  00  ;  --3;  —  i.  The  plane  T  =  00  ;  —  i;  —3.  Find 
the  plan  and  elevation  of  the  intersection  of  these  planes. 

131.  The  plane  S  =  o;  +60;  —60.  The  plane  T  =  00  ;  —4;  —3.  Find 
the  plan  and  elevation  of  the  planes  S  and  T. 

132.  The  plane  S  =  o;  —90;  —75.  The  plane  T  =  —2;  —75;  —60. 
Find  the  plan  and  elevation  of  the  intersections  of  S  and  T. 

133.  The  plane  R  =  —2;  —90;  —270.  The  plane  S  =  +1;  —90; 
+270.    Find  the  plan  and  elevation  of  planes  S  and  T. 

134.  The  plane  S  =  —2;  +45;  —60.  The  point  M  =  —  2;  —  i;  —5. 
The  point  N=4-2;— 4;— i.    Find  where  the  line  MN  pierces  the  plane  S. 

135.  The  plane  S  =  x;  —4;  —2.  The  point  M  =  —3;  —  i;  —4. 
The  point  N  =  o;  — 3;  —  i.    Find  where  MN  pierces  plane  S. 

136.  The  plane  S  =  o;  4-75**;  -30°.  The  line  MN  is  in  the  third 
quadrant  parallel  to  the  G.  L.  2"  from  H  and  V.  Find  where  the  line  MN 
pierces  plane  S. 

137.  The  point  M  =  o;  —  i;  —  2.  The  point  N  =  —3;  —  i;  —4.  The 
plane  S  =  —  3;  —60°;  —60".    Find  where  MN  pierces  plane  S. 

138.  The  plane  S  =  o;  +135*';  —105°.  The  point  O  =  — i;  —3;  —2. 
Find  the  shortest  distance  from  O  to  plane  S. 

139.  The  plane  R  =  00 ;  —3;  —2.  The  point  O  =  — o;  —3;  —4. 
Find  the  shortest  distance  from  O  to  the  plane  R  and  show  its  plan  and 
elevation. 

140.  The  plane  S  =  o;  +60°;  —45°.  Pass  a  plane  parallel  to  plane  S 
and  2"  from  it. 

141.  The  plane  S  =  o;  +75°;  —60°.  The  plane  R  =  —2;  +75;  —60. 
Find  the  shortest  distance  between  the  planes. 

142.  The  plane  S  =  o;  -45°;  +60°.  The  point  M  =  -2";  -2";  x. 
M  is  one  comer  of  a  regular  square  pyramid  whose  base  lies  in  plane  S.  The 
base  of  the  pyramid  is  a  2"  square  and  the  altitude  is  3".    Draw  a  plan, 


PROBLEMS  ON  POINTS,  LINES,  AND   PLANES  69 

elevation,  and  end  view  of  the  pyramid  when  one  edge  of  its  base  is  parallel 
to  sS. 

143.  The  plane  R  =  o;  +45°;  -60°.  The  point  0  is  +3'';  -y"]  -2", 
This  point  is  the  center  of  a  i"  square  which  lies  in  plane  R  with  one  edge 
making  an  angle  of  30  degrees  with  Rr.  Draw  the  plan,  elevation,  and  left 
end  view  of  a  cube,  one  of  whose  faces  is  this  square. 

144.  A  point  of  light  is  located  at  +3;  +6;  +3.  A  ray  of  light  from 
this  point  strikes  a  mirror  in  H  at  the  point  o;  +2;  o  and  is  reflected  to  V. 
Find  the  coordinates  of  the  point  where  the  reflected  ray  strikes  V.  (It 
will  be  remembered  that  the  angles  of  incidence  and  reflection  are  equal.) 

145.  A  ray  of  Ught  comes  from  the  point  +6;  +5;  +4,  and  strikes  a 
mirror  at  the  point  -f-3;  +3;  +2.  The  plane  of  the  mirror  inclines  60  de- 
grees to  V  and  is  perpendicular  to  H.     Find  where  the  reflected  ray  strikes  V. 

61.  Proposition  11.  Given  the  plan  and  elevation  of  a  line 
to  find  its  plan  and  elevation  upon  any  given  auxiliary  plane. 

Discussion.  The  projection  of  any  point  upon  any  plane  is 
the  foot  of  a  perpendicular  from  the  point  to  the  plane.  If 
perpendiculars  be  drawn  from  any  two  points  of  the  given  line 
to  the  given  auxiliary  plane  and  the  points  where  these  perpen- 
diculars pierce  this  plane  be  found,  these  piercing  points  when 
joined  will  be  the  required  projection  of  the  line. 

Construction.  The  problem  in  Fig.  56  is  to  find  the  plan  and 
elevation  of  the  face  MNOP  of  the  casting  on  the  plane  R  to 
which  this  face  is  parallel.  From  each  corner  of  the  face  draw  a 
perpendicular  to  the  plane  R  and  find  where  these  perpendiculars 
pierce  the  plane.  The  plan  view  of  the  projection  on  R  is  abed 
—  these  points  being  the  plan  view  of  the  piercing  points  of  the 
perpendiculars  —  and  the  elevation  is  a'b'c'd'.  AhBhChDh 
shows  the  face  in  its  true  size  after  it  has  been  revolved  into  H 
about  rR  as  an  axis,  and  set  to  one  side  of  the  actual  plane  of 
revolution  so  as  to  avoid  confusion  in  the  drawing. 

62.  Special  Cases.     To  project  a  line  upon  any  plane: 

1.  When  the  line  is  parallel  to  the  G.  L. 

2.  When  the  plane  is  parallel  to  the  G.  L. 

3.  When  the  plan  and  elevation  of  the  given  line  are  parallel 

to  the  corresponding  traces  of  the  given  plane. 

63.  Proposition  12.  Given  the  plan  and  elevation  of  a  Une 
and  of  a  point,  to  find  the  shortest  distance  between  them. 


70 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


Discussion:    First  Method.     If  the  line   and   the  point  be 
revolved  into  H  or  V  about  the  trace  of  the  plane  which  contains 

Ch 


Fig.  56. 

them  they  will  be  shown  in  this  revolved  position  in  their  true 
relation.  If  now  from  the  revolved  position  of  the  given  point 
a  perpendicular  be  drawn  to  the  revolved  position  of  the  line, 


PROBLEMS  ON  POINTS,  LINES,  AND   PLANES  71 

this  perpendicular  will  be  the  measure  of  the  shortest  distance 
from  the  point  to  the  given  line.  If  a  plan  and  elevation  of  this 
distance  be  required  the  point  must  be  revolved  back  to  the  first 
position,  and  with  it  the  perpendicular. 

Construction.  Let  the  construction  be  made  in  accordance 
with  the  discussion. 

Discussion:  Second  Method.  It  not  infrequently  happens 
that  the  relation  of  the  given  point  to  the  given  line  is  such  as 
to  make  the  construction  by  the  first  method  awkward.  In 
such  cases  it  will  be  found  more  convenient  to  pass  a  plane 
through  the  point  perpendicular  to  the  line,  and  by  finding  where 
the  given  line  pierces  this  plane  a  point  will  be  found  which,  when 
joined  to  the  given  point,  will  be  the  shortest  distance  from  the 
given  point  to  the  given  line. 

Construction.  Fig.  57  shows  such  a  case.  The  problem  is  to 
find  the  length  of  the  center  line  required  for  a  timber  to  reach 
from  point  A  to  the  second  timber  whose  center  line  is  OQ. 
Through  A  the  plane  R  is  passed  perpendicular  to  OQ.  (Article 
48.)  The  Hne  OQ  pierces  this  plane  at  B.  (Article  56.)  AB 
then  is  the  required  shortest  distance  and  abi  shows  its  true 
length. 

64.  Proposition  13.  Given  the  plans  and  elevations  of  two 
lines  not  in  the  same  plane  to  find  the  plan  and  elevation  of  the 
shortest  Hne  which  can  be  drawn  between  them. 

Discussion.  If  through  one  of  the  given  lines  a  plane  be 
passed  parallel  to  the  other,  and  if  upon  this  plane  the  second 
line  be  projected,  this  projection  will  intersect  the  first  line  at  a 
point  which  will  be  one  end  of  the  shortest  connecting  line.  If 
now  at  this  point  a  line  be  erected  perpendicular  to  the  plane  it 
will  intersect  the  second  line  at  a  point  which  will  be  the  other 
end  of  the  shortest  connecting  Hne.  If  the  length  of  this  Hne  be 
required  it  may  be  found  as  in  Article  38. 

Construction.  Fig.  58  shows  two  timbers  lying  in  different 
planes.  The  problem  is  to  find  the  shortest  third  timber  to 
join  them  —  center  Hnes  only  being  considered.  Through  PQ 
pass  a  plane  parallel  to  AB.  This  plane  is  rRr'.  Upon  plane 
R  project  AB.    (Article  61.)     This  projection  on  R  is  shown  by 


72 


ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 


Fig.  57. 


PROBLEMS  ON  POINTS,  LINES,   AND   PLANES 


73 


mn  and  m'n'.  Now  at  O,  where  PQ  and  MN  intersect,  erect  a 
perpendicular  to  the  plane  R.  This  perpendicular  intersects 
AB  at  C.  CO  then  is  the  center  line  of  the  shortest  timber  that 
can  connect  PQ  and  AB.  Its  true  length  may  be  found  as 
in  Article  38. 


Fig.  58. 


65.  Proposition  14.  Given  the  plan  and  elevation  of  a  line 
to  find  the  true  size  of  the  angle  it  makes  with  a  given  plane. 

Discussion.  The  angle  a  line  makes  with  a  plane  is  equal  to 
the  angle  a  line  makes  with  its  projection  on  that  plane.  If, 
therefore,  the  given  Une  be  projected  on  the  given  plane  and  the 
true  size  of  the  angle  between  the  given  line  and  the  projection 
on  the  given  plane  be  measured  it  will  be  the  required  angle  the 
given  line  makes  with  the  given  plane. 


74 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


Construction.     In  Fig.  59  the  problem  is  to  find  the  angle 
between  the  plane  S  and  the  center  Une  NQ  of  a  tunnel.    By 


Fig.  59. 

Article  56,  find  where  the  center  line  NQ  pierces  the  plane  S. 
This  point  is  M,  and  it  is  the  vertex  of  the  required  angle.  From 
any  other  point  N  of  NQ  draw  a  line  perpendicular  to  the  plane 


PROBLEMS   ON   POINTS,   LINES,   AND   PLANES 


75 


S  and  find  where  it  pierces  the  plane  S.  This  point  is  P, 
shown  in  plan  at  p  and  in  elevation  at  p',  and  MP  is  the 
projection  of  MN  upon  S.  NMP,  then,  is  the  angle  the  line 
NQ  makes  with  the  plane  S.  Its  true  size  is  shown  at  NhMhPh. 
(Article  50.) 

Further  Discussion.    In  case  only  the  size  of  the  angle  the  line 
makes  with  the  plane  is  required  a  shorter  method  may  be  used. 


Fig.  60. 

In  the  problem  just  solved  it  will  be  observed  that  since  the  angle 
MPN  is  a  right  angle  —  NP  being  perpendicular  to  MP  —  the 
angle  PNM  is  the  complement  of  the  required  angle.  If,  there- 
fore, from  any  point  of  the  given  line  a  perpendicular  be  drawn 
to  the  given  plane  the  angle  between  the  perpendicular  and  the 
given  line  will  be  the  complement  of  the  required  angle.  The 
true  size  of  this  complement  may  be  found  and  subtracted  from 
90  degrees  thus  giving  the  size  of  the  required  angle. 


76  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

Construction.  Fig.  60  shows  the  method  of  finding  the  size 
of  the  angle  between  the  plane  S  and  the  line  OQ.  The  perpen- 
dicular from  O  to  S  pierces  V  at  P;  OQ  pierces  V  at  Q.  r'R  is 
the  V  trace  of  the  plane  of  the  angle  QOP  and  its  true  size  may- 
be found  as  shown.  Subtracting  q'Ovp'  from  90  degrees  gives 
NOvp',  the  size  of  the  angle  OQ  makes  with  S. 

66.  Special  Cases,  i.  Find  the  angle  a  given  line  makes  with 
H,  V,  or  P. 

2.  Given  the  plan  and  elevation  of  a  Hne  and  of  its  projection 
on  some  plane  to  locate  the  traces  of  the  plane. 

PROBLEMS 

146.  The  plane  T  =  o; -j-60;  —  60.  The  point  M  =  o;  —  i;  — 1|. 
The  point  N  =  +2;  — i;  — i.  Find  the  plan  and  elevation  of  the  line  MN 
when  projected  upon  plane  T. 

147.  The  plane  S  =  00 ;  -3'';  +4''.  The  line  MN  is  parallel  to  the 
G.  L.  \"  from  H  and  V  in  the  third  quadrant.  Find  the  plan  and  elevation 
of  MN  when  projected  upon  plane  S. 

148.  The  point  M  =  o;  o;  o.  The  point  N  =  +3;  —3;  —3.  The 
point  O  =  +2;  —3;  —I.  Find  the  shortest  distance  from  O  to  MN  and 
show  its  plan  and  elevation. 

149.  A  tunnel  runs  from  a  point  N  =  o;  —  6;  o  S.  60°  W.  on  a  falling 
6  per  cent  grade.  The  bottom  of  a  shaft  is  located  at  a  point  —  3 ;  —  2 ;  — 10. 
Find  the  actual  length  of  the  shortest  tunnel  which  will  connect  the  bottom 
of  the  shaft  and  the  tunnel  and  show  its  plan  and  elevation. 

150.  A  drain  runs  N.  30°  E.  from  a  point  o;  o;  —  i  on  a  falling  10  per  cent 
grade.  Show  the  plan  and  elevation  and  find  the  length  of  the  shortest 
connection  between  the  drain  and  the  point  + 1 ;  —  2 ;  o. 

151.  A  steam  pipe  runs  through  the  points  M  =  o;  —3;  —  i  and 
N  =  +3;  —6;  —I.  A  second  pipe  runs  through  the  points  A  =  —  i;  —  i; 
—4  and  B  =  +2;  — i;  —2.  Find  the  shortest  connection  between  the 
pipes  and  give  its  true  length.     Use  center  lines  only. 

152.  A  tunnel  joins  the  bottoms  of  two  shafts  which  are  located  at 
o;  —8;  —50  and  +50;  —25;  —75.  A  second  tunnel  starts  at  o;  o;  —75 
and  runs  N.  45"  E.  on  a  rising  20  per  cent  grade.  Draw  a  plan  and  elevation 
of  the  shortest  connecting  tunnel  which  can  join  these  two  and  give  its 
actual  length. 

k53.  The  plane  of  an  ore  body  is  located  at  o;  +75°;  —60°.  This  plane 
is  cut  by  a  tunnel  at  the  point  O  =  —2;  —  3;  x.  The  tunnel  runs  S.  60°  E. 
through  this  point  on  a  rising  10  per  cent  grade.  Find  the  angle  the  timnel 
makes  with  the  plane  of  the  ore. 


PROBLEMS  ON  POINTS,  LINES,  AND   PLANES  77 

154.  The  plane  of  a  body  of  ore  is  located  at  S  =  o;  +75°;  —60**.  A 
shaft  is  sunk  at  a  point  +3;  —  3;  o.  Find  the  angle  the  shaft  will  make 
with  the  plane  of  the  ore  and  how  deep  the  shaft  will  have  to  be  in  order  to 
reach  the  ore. 

ft 5.  A  brace  14'  long  meets  the  ground  8'  in  front  of  a  wall  and  rests 
agi-inst  the  wall  6"  above  the  ground.  Find  the  angles  the  brace  makes 
with  the  ground  and  the  wall. 

67.  Proposition  15.  Given  the  angles  a  line  makes  with  H 
and  V,  and  the  plan  and  elevation  of  some  point  on  the  line,  to 
draw  the  plan  and  elevation  of  the  line. 

Discussion.  If  a  line  of  any  assimied  length  be  drawn  through 
the  point  parallel  to  H  and  making  the  required  angle  with  V  the 
plan  view  of  the  line  will  be  equal  in  length  to  the  plan  view  of 
the  required  Une.  Also,  if  a  line  be  drawn  through  the  given 
point  parallel  to  V  and  making  the  required  angle  with  H  the 
elevation  of  this  Une  will  be  equal  in  length  to  the  elevation  of 
the  required  line.  Having  the  length  of  the  two  views  and  the 
location  of  one  point  the  Une  may  be  constructed. 

Construction.  In  Fig.  61,  0  is  the  bottom  of  a  shaft  from 
which  runs  a  tunnel  with  a  given  grade  and  direction.  The 
grade  is  equivalent  to  the  angle  it  makes  with  H  and  its  direction 
is  equivalent  to  the  angle  it  makes  with  V. 

Draw  a  line  through  O  parallel  to  H  making  the  given  direc- 
tion with  V.  Assume  OQ  as  the  length  of  the  tunnel,  o'q", 
then,  is  the  length  of  the  elevation  of  the  required  line.  Likewise 
draw  through  O  a  line  parallel  to  V  and  making  the  given  grade 
with  H,  and  make  it  equal  in  length  to  OQ.  op',  then,  is  the 
length  of  the  plan  view  required.  Revolve  these  two  lines  until 
P  and  Q  coincide  at  N;  on  and  o'n'  then  will  be  the  required  plan 
and  elevation  of  the  center  line  of  the  tunnel. 

Note.  The  sum  of  the  angles  given  must  not  exceed  90 
degrees  or  the  problem  becomes  impossible. 

68.  Proposition  16.  Given  two  planes  to  find  the  angle  be- 
tween them. 

Discussion.  The  angle  between  the  planes  is  measured  in  a 
plane  perpendicular  to  both  planes  or  to  their  line  of  intersection. 
If,  then,  a  plane  be  passed  through  any  point  of  a  line  common 


78 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


to  both  planes  and  the  intersection  of  this  plane  with  both  of  the 
others  be  found  the  angle  thus  formed  will  be  the  required  angle 
between  the  given  planes,  and  its  size  may  be  determined. 

Construction.    In  Fig.  62  is  shown  a  plan  and  elevation  of  a 
trough.    The  problem  is  to  find  the  angle  between  its  sides. 


Fig.  61. 

Assume  the  point  0  on  the  line  common  to  the  planes  of  the  two 
sides,  and  through  this  point,  by  Article  48,  pass  a  plane  perpen- 
dicular to  the  Hne.  This  plane  is  rRr'.  Now  by  Article  54  find 
the  intersection  of  R  and  S,  and  R  and  T.  This  is  OP  in  the  first 
case  and  OQ  in  the  second,  and  POQ  is  the  required  angle.  Its 
true  size,  pOnq,  may  be  found  by  Article  50. 

69.  Special  Cases,     i.  Find  the  angle  a  plane  makes  with  H 
and  V. 


PROBLEMS  ON  POINTS,  LINES,   AND  PLANES 


79 


2.  Find  the  angle  between  two  planes  which  are  parallel  to 
the  G.  L. 

70.  Proposition  17.  Given  one  trace  of  a  plane  and  the  angle 
the  plane  makes  with  the  corresponding  plane  of  projection  to 
find  the  other  trace. 

/  Oh  / 

/   „.„„    / 


Fig.  62. 


Discussion.  The  vertex  of  the  given  angle  will  lie  on  the  given 
trace  and  the  angle  will  he  in  a  plane  perpendicular  to  the  given 
trace.     This  angle  may,  therefore,  be  constructed  and  since  one 


8o 


ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 


side  of  it  lies  in  the  required  plane  the  piercing  point  of  this  side 
will  lie  on  the  required  trace. 

Construction.    Let  the  problem  be,  Fig.  63,  to  find  the  plane 
of  the  ore  body  whose  strike  is  the  line  sS  (the  ^'strike"  is  the 


s 

/ 


Fig.  63. 

direction  of  any  horizontal  line  in  the  plane)  and  whose  dip  is 
60  degrees  (the  "dip"  is  the  inclination  of  the  plane  to  the  hori- 
zontal).   Lay  of!  an  angle  of  60  degrees  with  its  vertex  o  on  sS 


PROBLEMS  ON  POINTS,   LINES,   AND   PLANES 


8l 


and  one  side  OP  perpendicular  to  sS.  Pnop  will  be  the  revolved 
position  of  the  given  angle  the  ore  body  makes  with  H,  and  Ph 
will  be  the  revolved  position  of  the  point  where  OP  pierces  V. 
Since  the  line  OP  must  lie  in  the  plane  of  the  ore  revolve  the  angle 
till  it  is  perpendicular  to  H,  and  Ph  will  fall  at  a  point  shown  in 
plan  and  in  elevation  at  p  and  p'.  P,  then,  is  a  point  in  s'S, 
the  required  trace. 


71.  Special  Case.  Given  the  angles  a  plane  makes  with  H 
and  V  and  the  location  of  one  point  in  the  plane  to  find  the  traces 
of  the  plane. 

Discussion.  If  through  the  given  point  a  line  be  drawn 
perpendicular  to  the  required  plane  this  Une  will  make  with  H 
and  V  angles  equal  to  the  complements  of  the  angles  the  plane 
makes  with  H  and  V.  If,  therefore,  a  line  be  drawn  through  the 
given  point  making  with  H  and  V  complements  of  the  given 
angles,  this  line  will  be  perpendicular  to  the  required  plane,  and 
the  plane  may  then  be  located. 


82  ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 

Note.  The  sum  of  the  given  angles  must  not  be  less  than  90 
degrees  else  the  problem  will  be  impossible. 

Construction.  Let  0  be  the  given  point  in  Fig.  64  and  let  it 
be  required  to  find  the  traces  of  the  plane  through  0  which  makes 
60  degrees  with  H  and  45  degrees  with  V. 

By  Article  67,  draw  a  line  through  0  making  an  angle  of  30 
degrees  with  H  and  45  degrees  with  V.  Now,  by  Article  48,  pass 
through  0  a  plane  perpendicular  to  the  Hne.  sSs'  is  such  a  plane 
and  since  it  is  perpendicular  to  a  Une  making  30  degrees  with  H 
and  45  degrees  with  V  the  plane  itself  makes  60  degrees  with  H 
and  45  degrees  with  V.     S,  then,  is  the  required  plane. 

PROBLEMS 

156.  Through  the  point  —6;  —4;  —3  draw  a  line  3"  long  which  will 
make  angles  of  30  degrees  with  H  and  V. 

157.  The  bottom  of  a  shaft  is  located  at  o;  —6;  —10.  Through  this 
point  a  tunnel  is  driven  making  an  angle  of  60  degrees  with  V  and  15  degrees 
with  H.     Give  the  bearing  of  the  tunnel. 

158.  A  guy  wire  is  anchored  at  o;  —  3;  o  and  is  fastened  to  the  top  of  a 
stack  at  +3;  +1;  X.  Find  how  high  the  stack  is  when  the  wire  makes  an 
angle  of  60  degrees  with  the  ground.  Find  also  the  angle  the  wire  makes 
with  V. 

159.  Find  the  angle  between  the  end  and  side  of  a  floor  hopper;  size  of 
opening  in  floor  6'  by  8',  size  of  outlet  12"  by  12",  depth  of  hopper  3'. 

160.  Find  the  angles  between  the  sides  of  a  hood  which  fits  in  the  corner 
of  a  room.  The  opening  of  the  hood  is  24"  square  and  its  outlet  is  6"  square. 
The  distance  between  opening  and  outlet  is  18''. 

161.  Three  shafts  are  sunk  to  the  plane  of  an  ore  body.  One  is  located 
at  the  point  —3;  —4;  —10;  a  second  is  located  at  o;  —  i;  —18;  the  third 
is  located  at  +1;  —6;  —24.  Find  the  angle  the  plane  of  the  ore  makes 
with  H  and  V. 

162.  A  V-shaped  flume  runs  N.  30°  E.  on  a  falling  10  per  cent  grade.  The 
bottom  of  the  flume  runs  through  the  point  —  2;  o;  —2  and  the  angle  be- 
tween the  sides  of  the  flume  is  30  degrees.  Draw  a  plan  and  elevation  of 
the  flume  assuming  it  to  be  made  up  of  a  2"  by  10"  piece  fastened  to  a  2"  by 
8"  piece. 

163.  An  ore  body  dips  60  degrees  and  strikes  S.  15°  W.  from  the  point 
—  2;  —6;  —10.  Show  its  outcrop  and  find  the  angle  it  makes  with  the 
vertical  plane. 

164.  The  bottom  of  a  well  is  at  o;  —6;  —18.  The  plane  of  a  stratum 
of  clay  through  this  point  dips  60  degrees  and  makes  an  angle  of  75  degrees 
with  V.    Find  the  traces  of  this  plane. 


PROBLEMS  ON  POINTS,  LINES,  AND   PLANES  83 

165.  The  outcrop  of  a  stratum  of  stone  runs  N.  60°  W.  from  the  point 
4-6;  —4;  o  and  dips  15  degrees.     Find  the  angle  the  stratum  makes  with  V. 

166.  The  wedge-shaped  prow  of  a  snow  plough  makes  angles  of  60  de- 
grees with  H  and  V,  the  planes  of  the  faces  inclining  so  as  to  meet.  Find 
the  angle  at  which  a  plate  will  have  to  be  bent  in  order  to  form  a  shoe  over 
the  intersection  of  these  planes. 


CHAPTER  XI 
SURFACES 

72.  A  surface  is  the  area  made  by  a  line  moving  according  to 
some  law. 

In  descriptive  geometry  surfaces  are  classified  according  to  the 
character  of  the  moving  line,  called  the  generatrix,  and  the  law 
controlling  its  motion. 

73.  The  generatrix  may  be  either  a  straight  or  a  curved  line 
and  its  motion  may  be  controlled  in  one  of  several  ways.  It 
may  be  made  to  move  so  as  to  touch  other  lines  either  straight 
or  curved,  called  directrices,  and  to  remain  parallel  to  a  plane, 
called  a  plane  director;  or  it  may  be  made  to  revolve  about 
another  line,  called  the  axis  of  revolution;  or  its  motion  may  be 
controlled  in  other  ways.  All  surfaces,  however,  are  generated 
by  a  line  whose  motion  is  controlled  in  some  specified  way. 

74.  Surfaces  thus  generated  are  divided  into  two  general 
classes:  ruled  surfaces  and  double  curved  surfaces.  A  ruled  sur- 
face is  generated  by  the  motion  of  a  straight  Hne.  A  double 
curved  surface  is  generated  by  the  motion  of  a  curved  line.  It 
is  obvious,  therefore,  that  a  straight  edge  may  be  made  to  coin- 
cide with  a  ruled  surface  while  it  will  coincide  with  a  double 
curved  surface  in  one  point  only;  this  fact  offers  a  convenient 
method  of  identifying  surfaces. 

75.  There  are  three  kinds  of  ruled  surfaces:  Plane,  single 
curved,  and  warped. 

A  plane  surface  is  generated  by  a  line  moving  so  as  to  con- 
stantly touch  straight  line  directrices  which  either  intersect  or 
are  parallel.  Thus,  in  Fig.  65,  the  line  MN  is  the  generatrix, 
and  the  traces  of  the  plane,  sS  and  s'S,  are  the  intersecting  direc- 
trices. The  hne  MN,  then,  moves  so  as  to  constantly  touch 
these  two  lines,  thus  generating  the  plane  sSs'.  In  further  dis- 
cussions of  plane  surfaces  it  will  be  understood  that  surfaces 

84 


SURFACES 


ss 


with  plane  faces  are  meant;  the  intersections  of  the  plane  faces, 
or  planes,  form  the  edges  of  the  surfaces  which  hereafter  will  be 
called  plane  surfaces. 

Any  combination  of  intersecting  planes,  usually  three  or  more, 
forms  a  plane  surface.  Many  of  these  combinations  have 
geometrical  names  such  as  prisms,  pyramids,  etc.,  with  any 
number  of  faces  from  three  upward.  Some  of  these  surfaces 
have  commercial  names  which  identify  the  uses  to  which  they 
are  put,  as,  for  example,  the  hopper  which  geometrically  is  a 


Fig.  65.  Fig.  66. 

truncated  rectangular  pyramid.  It  is  with  such  surfaces  as  have 
commercial  uses  that  this  text  has  to  do  rather  than  with  those 
which  possess  mathematical  interest  only. 

76.  There  are  three  kinds  of  single  curved  surfaces:  cylin- 
drical, conical,  and  convolute.  All  of  these  surfaces  are  ruled 
and  the  rulings  —  or  the  positions  of  the  generating  line  —  are 
called  elements. 

A  cylindrical  surface  is  generated  by  a  straight  line  moving 
so  as  to  always  remain  parallel  to  its  first  position  and  to  con- 
stantly touch  a  plane  curve.^  When  this  plane  curve  is  closed 
the  surface  is  called  a  cylinder.     In  Fig.  66  the  generating  line 

^  A  plane  curve  is  one  which  lies  in  a  plane,  as  a  circle;  a  space  curve  is  one 
which  does  not  Ue  in  a  plane,  as  a  helix. 


86 


ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 


BX  moves  so  that  it  remains  parallel  to  its  first  position,  BX, 
and  constantly  touches  the  curve  ABC,  thus  generating  a  cylin- 
drical surface.    In  Fig.  67  the  directrix  is  taken  as  a  circle  in  H 

and  since  this  is  a  closed  curve 
the  resulting  surface  is  called  a 
cylinder.  It  will  be  noted  that 
the  elements  of  a  cyhnder  are 
parallel  to  each  other.  A  conical 
surface  is  a  surface  generated  by 
a  straight  line  so  moving  that 
one  point  remains  stationary 
while  the  line  constantly  moves 
along  a  plane  curve.  This  sta- 
tionary point  is  called  the  apex, 
and  it  is  obvious  that  since 
the  generating  line  extends  beyond  the  apex  there  will  be 
generated  two  surfaces,  called  nappes,  one  on  each  side  of  the 


Fig.  68. 


apex.     In  discussions  of  conical  surfaces  in  this  text  only  one 
nappe  will  be  considered,   however.     When   the  plane   curve 


SURFACES 


87 


Fig.  69. 


directrix  is  a  closed  curve  the  surface  is  called  a  cone.  In  Fig.  68 
the  generatrix  moves  so  that  it  goes  through  the  point  0,  the  apex, 
and  constantly  touches  the  curve 
XY,  thus  generating  a  conical  sur- 
face. In  Fig.  69  the  curved  direc- 
trix is  a  circle  and  the  resulting 
surface  is,  therefore,  a  cone. 

A  convolute  surface  is  generated 
by  a  straight  line  moving  so  as  to 
be  constantly  tangent  to  a  space 
curve.     In  Fig.  70  the  space  curve 

1-2-3 II     is     the     curvilinear 

directrix  to  which  the  elements  are 
constantly  tangent,  and  a  portion 
of   the   surface   is   shown  by  the 

positions   of   the   generating   line  at  i,   2 A,  3B, 7F.     The 

most  important  convolute  surface 
commercially  is  the  helical  con- 
volute, so-called  because  its  cur- 
vilinear directrix  is  a  helix,  and 
since  this  surface  is  a  special  case 
of  helicoid  it  will  be  considered 
in  detail  in  connection  with  that 
surface. 

77.  A  warped  surface  is  a  ruled 
surface  generated  by  a  straight 
line  so  moving  that  the  elements 
of  the  resulting  surface  neither 
intersect  nor  are  parallel.  Thus 
in  Fig.  71  the  generating  line  AX 
so  moves  that  the  elements  of  the 
resulting  surface  neither  intersect 
nor  are  parallel.  In  this  surface 
the  lines  AB  and  CD  are  recti- 
^^'  ^°'  linear  directrices  and  the  genera- 

trix, AX,  constantly  touches  them;   the  H  plane  is  the  plane 
director  as  all  of  the  positions  of  AX  remain  parallel  to  this  plane. 


88 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


There    are    innumerable    warped    surfaces    since    there   are 

innumerable  combinations 
of  generatrices,  directrices, 
and  plane  directors  but 
comparatively  few  of  them 
are  important  commer- 
cially, and  these  will  be 
discussed  in  detail  in  a 
later  chapter. 

78.  A  surface  of  resolu- 
tion is  generated  by  the 
motion  of  a  line  about 
another  line  called  the  axis 
of  revolution.  The  gen- 
erating line  may  be  either 
straight  or  curved ;  in  case 
the  generatrix  is  a  straight 


Fig.  71. 


line  the  resulting  surface  will  be  a  ruled  surface  of  revolution ;  if 
the  generatrix  is  a  curved  line 
the  resulting  surface  will  be  a 
double  curved  surface  of  revo- 
lution. In  Fig.  72  the  circle 
whose  center  is  at  O  is  re- 
volved about  the  axis  of 
revolution  AB .  The  resulting 
surface  is  a  double  curved 
surface  of  revolution  called  an 
annular  torus.  In  the  figure 
one-quarter  of  the  surface  has 
been  cut  away  in  order  to  show 
the  curvilinear  generatrix. 
Whether  the  generatrix  is 
straight  or  curved  all  surfaces 
of  revolution  have  a  circular 
section  in  planes  perpendic- 
ular to  the  axis  of  revolution. 

79.   Developable  surfaces  are  those  which  may  be  rolled  out 


SURFACES 


89 


on  a  plane  surface  so  that  all  the  elements  will  coincide  with  the 
plane.  In  Fig.  73  is  shown  a  cone  lying  in  a  plant  so  that  the 
element  OX  coincides  with  plane.  If  the  cone  be  now  rolled 
out  the  curve  of  the  base  will  rectify  itself  on  the  line  xx  as  the 
elements  of  the  surface  successively  come  in  contact  with  the 
plane.  The  result,  oxx^  is  called  a  development  of  the  cone. 
The  only  developable  surfaces  are  the  single  curved  surfaces  as 


these  obviously  are  the  only  surfaces  whose  elements  may  be 
made  to  lie  in  a  plane  without  changing  their  relation  to  each 
other.  In  practice,  however,  a  number  of  warped  and  double 
curved  surfaces  are  considered  developable;  this  is  made  possible 
by  approximate  methods  of  development  which  give  patterns 
which  conform  closely  enough  to  the  original  surface  for  practical 
purposes.  The  bottoms  of  hemispherical  water  tanks  are  good 
examples  of  this  fact. 


Surfaces  are 
either 


CLASSIFICATION  OF  SURFACES 

(Prisms 
Pyramids 
Polyhedrons 


Ruled.  Ruled 
Surfaces  may 
be 


Single  Curved  Surfaces  J  n?!!l^^'^ 
such  as 


2.  Double  Curved 


Warped  Surfaces  such 
as 


Double  Curved  Surfaces 
are  always  surfaces  of 
revolution  such  as 


•j  Cones 

[  Convolutes 

fCylindroids 
Conoids 
Helicoids 
etc. 

Spheres 

Ellipsoids 

Paraboloids 

Hyperboloids 

Tori 

etc. 


90 


ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 


Surfaces  of 
Revolution 
may  be 
either 


Ruled  Sur- 
faces such  as 


Single  curved  surfaces 
as 


2.  Warped  Surface  as 


Double  Curved  Sur- 
faces as 


Rt.  Cir. 
Cylinder 

Rt.  Cir. 

Cone 

The  hyperboloid 

of     revolution 

of  I  sheet. 

Only  case. 


Sphere 

Ellipsoid 

Paraboloid 
Hyperboloid  of  2  sheets 


/  Prolate 
^  Oblate 


CHAPTER  XII 
PLANE    SURFACES 

80.  A  plane  surface  is  represented  by  the  plan  and  elevation 
of  any  one  of  its  sections  and  of  its  edges.  While  the  drawing 
may  be  made  by  making  this  section  in  any  plane  it  is  usually 
taken  for  convenience  in  H  or  V;  in  whatever  plane  the  section 
lies  it  is  called  a  base. 

When  the  base  lies  in  H  or  V  it  will  be  apparent  that  it  is 
formed  by  the  intersections  of  the  traces  of  the  planes  of  the 
faces,  the  edges  of  the  base  in  reality  being  portions  of  the  traces. 
The  edges  of  the  surface  are  the  intersections  of  the  planes  of 
the  faces.  Thus,  in  Fig.  74,  the  planes  R,  S,  and  T  form  the 
faces  of  the  surface,  a  triangular  prism,  and  the  intersections  of 
these  planes,  XA,  YC,  ZB,  form  the  edges  of  the  surface;  the 
triangle  XYZ,  made  by  the  intersections  of  the  V  traces  of  these 
planes,  forms  the  V  base;  the  triangle  ABC  made  by  the  inter- 
sections of  the  H  traces  forms  the  H  base. 

81.  To  assume  a  point  on  a  plane  surface  any  line  as  MN, 
Fig.  74,  may  be  assumed  in  the  plane  R,  S,  or  T  by  Article  25, 
then  any  point  as  0  on  this  Hne  will  lie  on  the  surface  of  the 
prism. 

82.  In  case  the  given  base  of  the  surface  does  not  lie  in  H  or 
V  the  problem  may  be  solved  either  by  extending  the  surface 
so  as  to  obtain  a  base  in  either  H  or  V,  or  in  a  plane  parallel  to 
H  or  V,  or  by  using  the  plane  of  the  given  base  as  an  auxiUary 
plane  of  projection.  As  a  rule  the  problem  may  be  solved  more 
readily  by  finding  a  base  in  H  or  V. 

^  Note.  A  section  is  the  intersection  obtained  by  cutting  a 
surface  by  a  plane.  If  the  plane  of  the  section  is  perpendicular 
to  the  axis  of  the  surface  the  section  is  called  a  right  section; 
other  sections  are,  in  general,  oblique  sections. 

91 


92 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


83.  Proposition  18.  Given  the  size  and  location  of  the  right 
section  of  any  plane  surface  to  find  the  plan  and  elevation  of  the 
surface. 


Fig.  74. 


Discussion.  Since  the  right  section  Ues  in  a  plane  perpendicu- 
lar to  the  axis  the  edges  of  the  surface  will  be  perpendicular  to 
this  plane.  If,  then,  through  the  corners  of  the  given  right  sec- 
tion lines  be  drawn  perpendicular  to  the  plane  of  this  section 
these  lines  will  be  the  edges  of  the  surface  and  the  points  where 


PLANE  SURFACES 


93 


these  edges  pierce  H  and  V  will  determine  the  comers  of  the 
bases  in  H  and  V. 


Fig.  75. 


Construction.  In  Fig.  75,  T  is  the  plane  of  the  given  right 
section  whose  center  is  at  the  given  point  O.  The  section  is  to 
be  a  square  of  given  dimensions.     To  find  the  plan  and  elevation 


94  '         ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

of  this  section  O  is  revolved  into  V  at  Oy)  around  this  position 
of  O  the  square  may  be  constructed  in  its  true  size  and  revolved 
back  to  its  true  position  in  plane  T.  (Article  37.)  Through 
the  corners  of  the  square  in  its  true  position  draw  the  lines  which 
form  the  edges  of  the  surface.  Their  plans  and  elevations  will 
be  perpendicular  to  the  corresponding  traces  of  the  plane  T,  as 
ab  and  a'b'.  Extend  these  lines  until  they  pierce  H,  —  in  this 
case,  —  and  join  the  piercing  points.  The  area  thus  enclosed 
will  be  the  base  of  the  surface  in  H.  In  the  figure  this  area  has 
been  cross  hatched,  thus  representing  the  object  to  be  a  solid. 

84.  Proposition  19.  Given  the  plan  and  elevation  of  a  plane 
surface  to  find  the  true  shape  and  size  of  its  right  section. 

Discussion.  Pass  a  plane  perpendicular  to  the  axis  of  the 
surface;  this  will  be  the  plane  of  the  right  section.  Find  where 
the  edges  of  the  surface  pierce  this  plane;  these  points  when 
joined  will  determine  the  right  section.  Revolve  this  area  into 
H  or  V  about  the  H  or  V  trace  of  the  section  plane;  this  will 
give  the  true  shape  and  size  of  the  required  section. 

Construction.  The  figure  is  to  be  drawn  according  to  the 
above  discussion. 

85.  Proposition  20.  To  find  the  section  cut  from  any  plane 
surface  by  any  plane. 

Discussion :  First  Method.  Find  the  intersection  of  the  plane 
of  each  face  of  the  surface  with  the  given  cutting  plane.  Since 
each  of  the  lines  thus  found  is  common  to  both  the  surface  and 
the  cutting  plane  the  area  enclosed  by  these  intersections  will 
be  the  section  cut  from  the  surface  by  the  plane. 

Construction.  In  Fig.  76  the  surface  is  assumed  to  be  a 
rectangular  pyramid  cut  by  the  plane  sSs'.  The  plane  R  is  the 
left  face  of  the  solid  and  cuts  the  plane  S  in  the  line  AB ;  also  the 
plane  T,  which  is  the  plane  of  the  right  face  of  the  solid,  inter- 
sects the  plane  S  in  the  line  CD.  In  like  manner  the  fines  AD 
and  BC  may  be  found.  Since  these  four  fines  are  common  to 
both  the  surface  and  the  plane  S  the  area  they  enclose  will  be  the 
section  cut  from  the  pyramid  by  the  cutting  plane  S.  If  the  true 
size  of  this  area  is  desired  it  may  be  found  by  revolving  ABCD 
into  H  or  V  about  the  corresponding  trace  Ss  or  Ss'. 


PLANE   SURFACES 


95 


Discussion:  Second  Method.  Find  where  each  edge  of  the 
surface  pierces  the  cutting  plane.  Since  these  points  all  lie  in 
one  plane  and  are  common  to  the  surface  and  the  cutting  plane 
they  will,  when  joined,  form  the  required  intersection. 


Fig.  76. 

Construction.  In  Fig.  77,  O  is  the  apex  of  a  solid  octagonal 
pyramid  whose  base  is  in  a  plane  parallel  to  H.  By  Article  56 
find  where  each  edge  pierces  the  cutting  plane  S,  as  at  X,  Y,  etc. 
Join  these  points  in  order  and  the  area  thus  enclosed  will  be  the 
section  cut  from  the  surface  by  the  plane  S.     If  the  true  size  of 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


the  section  is  required  it  may  be  found  by  revolving  it  into  H  or 
V.  In  this  figure  the  section  is  revolved  into  H  about  the  trace 
sS  (Article  36),  and  its  true  size  is  shown  by  the  hatched  area 
at  XhYh,  etc. 


Fig.  77. 

86.  Proposition  21.    To  develop  any  plane  surface. 

Discussion.  Find  the  true  size  of  each  face  of  the  surface  and 
lay  these  faces  out  on  a  plane  with  their  common  edges  coin- 
ciding. 


PLANE   SURFACES 


97 


Construction.  In  Fig.  78  is  shown  a  hood  such  as  might  be 
used  over  a  forge.  To  make  a  development  of  it  first  find  the 
true  size  of  the  edge  OA.  At  any  convenient  place  on  the  paper 
lay  off  OA  equal  to  a'o",  then  with  A  as  a  center  and  AD  as  a 
radius  strike  an  arc  and  with  O  as  a  center  and  OD  equal  to  OA 
as  a  radius  strike  a  second  arc  intersecting  the  first  at  D.  The 
area  AOD  is  the  true  size  of  the  back  of  the  hood.  In  similar 
fashion  using  OD  as  a  common  edge  construct  the  area  OCD 


Fig.  78. 

equal  to  the  side  of  the  hood,  and  continuing  this  process  each 
face  of  the  hood  may  be  laid  out.  The  lips  of  the  hood,  or  the 
portions  lying  below  the  line  ABCD  are  laid  off  by  constructing 
rectangles  with  one  side  equal  to  AD,  DC,  CB,  and  AB  and  the 
other  side  equal  to  the  depth  of  the  lip.  These  rectangles  are 
of  course  joined  to  the  hood  along  the  lines  AD,  DC,  etc. 

In  case  the  edges  of  the  surface  to  be  developed  do  not  meet 
in  a  point,  as  in  the  case  of  a  pyramid,  or  are  not  parallel,  as  in 
the  case  of  a  prism,  the  development  may  be  made  as  shown  in 
Fig.  79.  The  method  used  is  the  same  in  principle  as  that  used 
in  Fig.  78,  modified  slightly  to  suit  the  altered  conditions. 

Construction.  Lay  off  the  area  BCHG  making  BC  =  be, 
CH  =  c'h^  BG  =  b'g'  and  GH  =  gh.     This  will  be  the  back  of 


98 


ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 


the  piece.  Now  with  C  as  a  center  and  a  radius  equal  to  c'e' 
strike  an  arc,  and  with  H  as  a  center  and  h'e'  as  a  radius  strike 
a  second  arc  intersecting  the  first  at  E.  This  point  E  is  the 
position  of  E  in  the  development  of  the  face  CDEH.  Now  to 
find  the  position  of  D  strike  arcs  with  C  as  a  center  and  c'd'  as 


Fig.  79. 


a  radius,  and  with  E  as  a  center  and  c'd'  as  a  radius.  These  arcs 
intersect  at  D  in  the  figure  and  thus  the  side  CDEH  may  be 
developed  by  dividing  it  into  triangles  and  locating  the  corners 
as  shown.  The  layout  for  the  remaining  portion  of  the  pieces 
will  be  apparent  from  an  inspection  of  the  figure. 


PLANE  SURFACES  99 

PROBLEMS  ON  PLANE  SURFACES 

167.  The  center  line  of  a  timber  12"  square  pierces  H  8''  in  front  of  V 
and  inclines  45  degrees  to  H  and  30  degrees  to  V.  Draw  a  plan,  elevation, 
and  end  view  of  the  timber, 

168.  The  center  line  of  a  timber  6"  square  pierces  H  at  O  (o";  —6";  o") 
and  pierces  V  at  Q  (18";  o'';  —12").  Draw  a  plan  and  elevation  of  the 
timber  and  show  its  intersection  with  H  and  V. 

169.  The  point  0  is  o";  -6";  o".  The  point  Q  is  -8'';  o";  -8".  The 
line  OQ  is  the  center  Une  of  a  rectangular  pipe  4"  by  6"  in  section.  Draw  a 
plan  and  elevation  of  the  pipe  and  find  the  true  size  of  the  hole  it  makes  in 
passing  through  the  profile  plane  located  at  —  $";  90;  90. 

170.  The  point  O  is  - 2";  -4";  o" .  The  point  Q  is  2";  o";  -4".  The 
line  OQ  is  the  center  Hne  of  an  hexagonal  prism  whose  base  in  H  is  a  regular 
hexagon  with  sides  1"  long.  Draw  a  plan,  elevation,  and  end  view  of  the 
prism  and  show  its  base  in  V. 

171.  Make  a  drawing  showing  how  to  trim  the  ends  of  a  2"  by  4"  stick 
so  that  it  will  fit  against  V  making  an  angle  of  45  degrees,  will  fit  against  H 
making  an  angle  of  30  degrees,  and  will  have  a  center  line  18"  long  between 
these  points. 

172.  The  plane  T  is  o";  —  60;  -f^o.  The  point  O  in  this  plane  is  —$"\ 
x"\  —3"  and  is  the  center  of  a  regular  hexagon  with  1"  sides  which  lie  in 
plane  T.  Draw  a  plan  and  elevation  of  the  prism  perpendicular  to  V  which 
could  cut  this  hexagon  from  the  plane. 

173.  The  point  O  is  o'';  -3";  -3".  The  point  Q  is  6";  -i";o". 
The  line  OQ  is  the  center  line  of  an  hexagonal  prism  whose  base  in  P  is  a 
regular  hexagon  with  1"  sides  and  center  at  0.  Draw  a  plan,  elevation, 
and  end  view  of  this  prism  and  find  the  true  shape  and  size  of  its  right  section. 

174.  The  base  of  a  pyramid  is  a  hexagon  with  i\"  sides  in  H  and  its 
center  at  the  point  o";  —3")  o".    The  apex  of  the  pyramid  is  at  Q  (— 4''; 

—  i";  —3")'  Draw  a  plan  and  elevation  of  this  pyramid  and  find  its  right 
section  at  a  point  2"  from  the  apex. 

175.  The  plane  T  is  o";  —60;  +45.    The  point  O  is  in  this  plane  at 

—  4";  —2";  x".  This  point  O  is  the  center  of  a  rectangle  2"  by  2,"  whose 
sides  incline  at  45  degrees  to  the  H  trace  of  plane  T.  The  rectangle  is  the 
base  of  a  quadrilateral  pyramid  whose  altitude  is  2".  Show  three  views 
of  the  pyramid. 

176.  A  trough  is  built  up  of  a  2"  by  2>"  piece  nailed  to  two  2"  by  6" 
pieces.  Draw  the  plan,  elevation,  and  end  view  of  a  section  18''  long  which 
slopes  30  degrees  to  H  and  V. 

177.  Make  a  development  of  a  hopper  whose  opening  is  a  rectangle  18" 
by  36''  and  whose  12"  by  12"  outlet  is  formed  by  cutting  the  hopper  by  a 
plane  inclining  30  degrees  to  the  horizontal.  The  distance  between  the 
centers  of  the  openings  is  18". 


lOO  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

178.  Make  the  pattern  for  a  chute  12"  square  in  section  to  connect  the 
opening  in  a  bin  wall  48''  above  the  floor  with  an  opening  in  the  floor  36" 
from  the  bin  wall. 

179.  An  air  duct  of  6"  by  8"  pipe  follows  the  line  of  intersection  of  the 
walls  and  ceiling  of  a  room  with  its  8"  side  against  the  ceiling.  Lay  out 
patterns  for  two-piece  elbow  needed  in  the  corners. 

180.  Lay  out  the  pattern  for  a  pipe  to  connect  a  12"  square  hole  in  a 
floor  with  a  bin  wall.  The  center  hne  of  the  pipe  inchnes  30  degrees  to  the 
floor,  45  degrees  to  the  bin  wall,  and  is  72"  long. 

181.  Design  the  transition  piece  to  connect  a  12''  square  air  duct  with  a 
4"  square  air  duct.  The  two  ducts  lie  on  the  floor  of  a  building  against  the 
wall  with  their  ends  6"  apart. 

182.  Design  a  ventilating  hood  for  the  comer  of  a  room.  The  opening 
of  the  hood  is  to  be  18"  by  24'',  the  outlet  is  to  be  in  the  comer  into  a  9"  by 
12''  pipe,  and  the  hood  is  to  be  16"  deep  with  a!'  lips. 

183.  A  12"  square  ventilating  pipe  runs  along  the  intersection  of  the 
ceiling  and  front  wall  of  a  room.  Design  the  elbow  which  will  take  the  pipe 
down  the  side  waU  from  the  comer  at  an  angle  of  30  degrees  with  the  ceiling. 

184.  Make  the  pattems  for  an  elbow  turning  an  angle  of  105  degrees 
using  4"  by  6"  pipe  with  the  plane  of  the  angle  in  the  direction  of  the  long 
dimensions  of  the  pipe. 

185.  Make  the  pattems  for  an  elbow  tuming  an  angle  of  60  degrees 
using  4"  by  6"  pipe  with  the  plane  of  the  angle  in  the  direction  of  the  short 
dimension  of  the  pipe. 

186.  A  water  duct  runs  down  the  side  of  a  building  at  an  angle  of  15  de- 
grees with  the  horizontal  and  tums  around  the  corner  of  the  building.  Lay 
out  pattems  for  an  elbow  to  make  the  turn  using  3"  square  pipe. 


CHAPTER  XIII 
CYLINDRICAL   SURFACES 

87.  A  cylindrical  surface  is  represented  by  the  plan  and 
elevation  of  some  curve  of  the  surface  and  of  the  Umiting  elements 
of  the  surface.  For  convenience  this  curve  is  usually  taken  in 
H  or  V  and  is  therefore  the  locus  of  the  points  in  which  the 
elements  of  the  surface  pierce  H  or  V. 

88.  In  a  cylinder  this  curve  is  called  a  base  and  for  conven- 
ience it  is  taken  as  a  rule  either  in  H  or  V,  or  in  a  plane  parallel 
to  H  or  V.  Since  the  base  of  a  cyUnder  must  be  a  closed  curve, 
cylinders  will  be  either  circular  or  elliptical  in  section,  or  the 
section  will  be  some  irregular  closed  plane  curve  such  as,  for 
example,  would  be  formed  by  cutting  a  corrugated  tin  pipe. 
When  a  cyUnder  with  a  circular  section  is  inclined  to  H  or  V  its 
bases  in  those  planes  will  be  ellipses;  when  the  cyhnder  is  elliptic 
cal  in  section  and  is  inclined  to  H  or  V  its  bases  will,  in  general, 
be  ellipses,  although  if  the  angle  of  the  incHnation  be  properly 
adjusted  they  may  become  circles.  Thus,  in  Fig.  80  is  a  cyUn- 
der with  a  circular  base  in  H.  This  cylinder  must  be  elliptical 
in  section.  In  Fig.  81  is  a  cyUnder  with  an  elUptical  base  in  V. 
This  cyUnder  may  be  either  circular  or  elUptical  in  section, 
according  to  the  angle  its  axis  makes  with  V. 

89.  CyUnders  are  defined  by  their  bases.  Thus:  a  cyUnder 
with  a  circular  base;  a  cylinder  with  an  elliptical  base,  etc.  A 
right  cylinder  is  one  whose  elements  are  perpendicular  to  the 
plane  of  its  base.  Such  cylinders  may  be  either  circular  or 
elliptical  in  section  and  are  generally  named  accordingly,  as  a 
right  circular  cyUnder,  or  a  right  elUptical  cyUnder. 

90.  To  assume  a  point  on  the  surface  of  a  cylinder  it  is  neces- 
sary only  to  assume  an  element  of  the  surface  and  on  this  element 
locate  the  point.  In  Fig.  80,  OP  is  the  axis  of  a  cyUnder  whose 
base  is  a  circle  in  H.     Assume  the  element  XY  parallel,  of  course, 


I02 


ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 


to  the  axis  and  piercing  H  in  the  base  at  Y.  Any  point  on  this 
element  will  be  a  point  on  the  surface  of  the  cylinder.  The 
element  projected  at  x'y'  has  two  plan  views,  one  at  xy  and  the 
other  at  qr.  Either  of  these  is  correct  as  one  is  on  one  side  of 
the  cylinder  and  the  other  is  directly  opposite  on  the  other  side. 
This  same  thing  is  shown  in  Fig.  8i.     If  the  point  Q  were  given 


Fig.  8o. 


Fig.  8i. 


by  its  plan  view  only,  two  V  projections  would  be  possible,  one 
at  q'  and  one  at  x',  since  the  element  through  Q  has  two  possible 
V  projections.  In  given  problems  it  will  always  be  stated  on 
which  element  to  take  the  point. 

91.  Proposition  22.  Given  the  axis  and  right  section  of  any 
cylinder  to  find  its  plan  and  elevation. 

Discussion:  First  Method.  Revolve  the  axis  parallel  to  H 
or  V.  In  this  position  draw  the  right  section.  Through  the 
ends  of  the  axes  of  the  curve  of  the  right  section  draw  elements; 
these  limiting  elements  will  pierce  H  or  V  at  the  ends  of  the  axes 
of  the  base. 

Construction.  In  Fig.  82,  OQ  is  the  given  axis  and  the  given 
right  section  is  a  circle  whose  diameter  is  dc  long.     Revolve  OQ 


CYLINDRICAL   SURFACES 


103 


parallel  to  V;  in  this  position  it  is  projected  at  oq^  and  o'q". 
At  any  convenient  point  P  place  the  given  right  section.  Its 
elevation  will  be  a  straight  line,  a"b",  since  the  axis  is  parallel  to 
V;  and  its  plan  view  will  be  an  ellipse  whose  major  axis  will  be 


d^c^  =  a"b''  and  whose  minor  axis  will  be  a^b^,  or  the  projection 
of  a'^b''  on  H.  Through  A  and  B  draw  Unes  parallel  to  OQ  in 
its  revolved  position;  these  will  be  limiting  elements  and  will 
pierce  H  at  y^  and  x^,  the  ends  of  the  major  axis  of  the  base  in  H. 
Likewise  draw  the  Hmiting  elements  through  D  and  C;  these 


I04  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

pierce  H  at  n^  and  m^  respectively  and  locate  the  ends  of  the 
minor  axis  of  the  base  in  H.  Now  revolve  the  axis  back  to  its 
given  position  OQ  and  locate  the  true  position  of  the  axes  of  the 
base.  Since  the  angle  that  OQ  makes  with  H  remains  constant 
the  axes  of  the  base  remain  constant  and  in  the  same  relation  to 
OQ.  Therefore,  x^  will  revolve  to  x,  y,  to  y,  n^  to  n,  and  m, 
to  m.  Construct  the  ellipse  of  the  base  and  draw  the  projec- 
tions of  the  limiting  elements. 

Discussion:  Second  Method.  Pass  a  plane  perpendicular  to 
the  given  axis;  this  plane  contains  the  given  right  section.  Con- 
struct the  projections  of  the  given  right  sections,  draw  the 
hmiting  elements,  and  find  where  they  pierce  H  or  V.  These 
points  will  give  the  axes  of  the  base  in  H  or  V. 

Construction.  Through  any  convenient  point  P,  Fig.  83,  pass 
a  plane  T  perpendicular  to  the  given  axis  OQ.  (Article  48.) 
Revolve  the  point  P  into  H  about  Tt.  With  this  point  P  as  a 
center  draw  the  given  right  section  in  its  true  size.  When  this 
circle  is  revolved  to  its  proper  position  in  plane  T  it  will  be  pro- 
jected as  an  ellipse  and  the  diameters  AC  and  BD  will  become 
the  major  and  minor  axes  of  this  elUpse  as  it  is  projected  in  H. 
Find  the  projections  of  A,  B,  C,  and  D  by  Article  37  and  through 
these  points  draw  the  projections  of  elements  of  the  surface. 
These  elements  pierce  H  at  yx  and  mn,  which  points  are  re- 
spectively the  ends  of  the  major  and  minor  axes  of  the  base. 
Draw  the  ellipse  determined  by  these  points  and  find  the  pro- 
jections of  the  hmiting  elements. 

92.  Proposition  23.  To  find  the  curve  cut  from  any  given 
cylinder  by  any  given  plane. 

Discussion.  Pass  planes  through  the  cyKnder  parallel  to  its 
axis.  These  auxihary  planes  cut  elements  from  the  surface  of 
the  cyhnder  and  Unes  from  the  cutting  plane.  These  fines  inter- 
sect the  elements  in  points  common  to  both  the  cyhnder  and  the 
cutting  plane.  If  a  sufficient  number  of  auxihary  planes  be 
used  enough  planes  will  be  obtained  to  plot  the  curve  of  inter- 
section. 

Construction.  When  the  cyhnder  is  perpendicular  to  H  or 
V.     In  Fig.  84  the  given  circular  cyhnder  with  its  base  in  H  is 


CYLINDRICAL   SURFACES 


los 


cut  by  a  plane  T.  Through  the  axis  pass  auxiliary  plane  R. 
This  plane  cuts  two  elements  from  the  cylinder  at  D  and  C,  and 
a  line  DC  from  the  plane  T.     This  hne  intersects  the  two  ele- 


ments at  D  and  C,  two  points  of  the  required  curve.  In  like 
manner  the  plane  U  cuts  two  elements  at  A  and  B  and  a  line  AB 
from  the  plane  T.     This  line  intersects  the  two  elements  at  A 


io6 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


and  B.    From  an  inspection  of  the  drawing  it  will  be  seen  that 
C  is  the  point  where  the  plane  T  enters  the  cylinder  and  the 


Fig.  84. 


point  D  is  where  the  plane  leaves  the  cylinder;  therefore,  C  and 
D  are  the  two  points  on  the  curve  of  intersection  farthest  apart, 
or  the  ends  of  the  major  axis  of  the  curve.     Since  the  auxiliary 


CYLINDRICAL  SURFACES  107 

plane  U  contains  the  line  AB  and  is  perpendicular  to  the  plane  R 
the  two  points  A  and  B  will  be  the  ends  of  the  minor  axis  of  the 
curve  of  intersection.  Now  if  AB  and  CD  be  revolved  into  V 
about  Tt',  the  true  size  of  the  curve  of  intersection  may  be 
found  as  shown.  If  the  projections  of  this  curve  of  intersection 
be  required  they  may  be  found  by  revolving  the  ellipse  AyBvCvDv 
to  its  original  position,  or  by  passing  more  auxiliary  planes 
parallel  to  U,  as  W,  and  locating  points  as  in  plane  U. 

Construction.  When  the  cylinder  is  oblique  to  H  or  V.  In 
Fig.  85  is  shown  a  cyHnder  oblique  to  H  with  circular  base  in  H. 
This  cylinder  is  cut  by  the  plane  T  as  in  the  case  just  discussed. 
Pass  planes,  such  as  R,  U,  W,  through  the  cylinder  parallel  to 
its  axis.  Each  plane,  except  the  plane  tangent  along  the  limit- 
ing  elements,  cuts  two  elements  from  the  cylinder  and  a  straight 
line  from  the  cutting  plane  T.  These  lines  are  of  course  parallel 
as  shown  at  AM,  BN,  CQ,  and  intersect  the  elements  of  the 
cylinder  in  points  of  the  curve  of  intersection,  as  A,  B,  C,  and 
D.  If  a  number  of  auxiliary  planes  are  passed  a  sufficient  num- 
ber of  points  will  be  determined  to  plot  the  curve  of  intersec- 
tion. In  finding  the  true  size  of  this  curve  the  points  may  be 
revolved  into  V  or  H  about  the  corresponding  trace.  In  this 
case  the  curve  is  revolved  into  V  about  the  V  trace  taking  the 
position  AyBvCyDv. 

93.  It  must  be  remembered  in  working  problems  of  this 
character  that  the  more  points  there  are  found  on  the  curve 
the  more  accurate  will  be  the  final  result.  In  the  figures  given 
as  illustrations  only  a  few  points  have  been  located  in  order  to 
avoid  confusion  in  the  drawings. 

It  must  also  be  remembered  that  only  by  solving  the  problem 
in  an  orderly  fashion  can  accurate  results  be  secured.  It  is  not 
wise  to  attempt  short  cuts,  and  in  the  end  it  will  be  found  most 
satisfactory  to  pass  the  auxiliary  planes  one  at  a  time  and  to 
determine  all  of  the  points  in  each  plane  before  passing  to  the 
next.  Any  attempt  to  pass  all  of  the  auxiliary  planes,  then  to 
draw  all  of  the  elements  these  planes  cut  from  the  cylinder,  and 
then  to  locate  all  of  the  lines  cut  from  the  cutting  planes  results 
either  in  throwing  the  problem  into  a  hopeless  confusion  of  lines. 


Io8  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


Fig.  8s. 


CYLINDRICAL  SURFACES 


109 


Fig.  86. 


no  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

or  else  requires  more  painstaking  work  to  avoid  this  tangle  than 
the  method  suggested.  This  condition  is  true  of  all  problems 
of  this  character. 

94.  Proposition  24.  Given  any  cylinder  to  find  its  right 
section. 

Discussion.  Pass  a  plane  through  the  cylinder  perpendicular 
to  its  axis  and  find  the  curve  of  intersection;  this  will  be  a  right 
section  of  the  cylinder. 

Construction.  If  both  the  true  size  and  the  projections  of 
the  right  section  are  required  the  problem  may  be  best  solved 
by  Proposition  23.  If,  however,  only  the  true  size  of  the  right 
section  is  required  the  solution  in  Fig.  86  will  be  found  more 
simple.  Let  OQ  be  the  axis  of  the  given  cylinder  and  T  the 
perpendicular  plane  containing  the  right  section.  The  axis  OQ 
pierces  the  plane  T  (Article  56)  at  P;  this  will  be  the  center  of  the 
right  section;  Now  draw  a  line  AB  through  P  parallel  to  the  H 
trace  of  T;  this  Hne  will  be  the  longest  line  which  can  be  drawn 
between  two  elements  of  the  surface;  hence  it 
is  the  major  axis  of  the  curve  of  intersection. 
To  find  the  minor  axis  draw  CD  through  P 
perpendicular  to  the  trace  tT.  Revolve  AB  and 
CD  into  V  about  the  V  trace  to  AyBy  and 
CyDy;  with  these  lines  as  axes  construct  the 
ellipse  which  will  be  the  right  section  of  the 
cylinder. 
95.  It  will  have  been  noted  in  connection  with 
Fig  87  Article    92    that   the    intersection    of    a    right 

cylinder  with  an  obhque  plane  is  a  curve  which 
is  symmetrical  about  its  center  Hues.  This  fact  is  utilized  in 
constructing  cylindrical  elbows  so  that  the  pipe  section  may 
remain  uniform  while  the  elbow  turns  an  angle. 

In  Fig.  87  is  shown  the  plan  and  elevation  of  a  cy Under  cut  by 
an  oblique  plane.  Since  this  curve  of  intersection  is  symmetri- 
cal the  upper  part  of  the  cylinder  may  be  turned  through  180 
degrees  to  the  position  shown  in  Fig.  88  and  the  two  parts  of  the 
cylinder  will  still  join  on  the  same  curve  of  intersection  while  the 
axis  turns  an  angle  of  90  degrees.     Fig.  89  shows  how  this  same 


CYLINDRICAL   SURFACES 


III 


Fig.  88. 


principle  may  be  applied  to  elbows  with  more  than  one  section. 
If  the  problem  be  to  draw  a  four  section  elbow  turning  an  angle 
of  6  degrees,  Fig.  90,  draw  two  lines 
making  the  given  angle  with  each  other, 
as  AO  and  DO,  and  then  divide  the  angle 
by  drawing  BO  and  CO  so  that  the  angle 

e 

BOA=-- 

3 

Find  the  points  i,  2,  3,  and  4  all  equi- 
distant from  O  and  with  these  points  as 
centers  lay  ofif  on  AO,  BO,  CO,  and  DO  the  diameters  of  the 

right  sections  of  each  part  of  the 
elbow.  The  limiting  elements 
may  be  drawn  through  the  ends  of 
the  diameter  of  the  right  section. 
96.  Proposition  25.  To  de- 
velop any  cyhnder. 
Discussion.  Since  the  ele- 
ments of  a  cyhnder  are  perpen- 
dicular to  the  plane  of  the  right 
section  the  curve  of  the  right  sec- 
tion will  roll  out  in  the  develop- 
ment into  a  straight  Hne  to  which 
all  the  elements  of  the  surface  are 
perpendicular.  Therefore,  lay  ofT 
a  line  equal  in  length  to  the  right 
section  and  erect  perpendiculars 
to  this  Hne.  If  the  distances  be- 
tween these  perpendiculars  be 
made  to  equal  the  distances  be- 
tween the  elements  along  the  curve 
of  the  right  section  and  the  perpen- 
diculars be  made  equal  in  length 
to  the  elements  the  result  will  be 
^^^-  ^9-  a  development  of  the  cylinder. 
Construction.  In  Fig.  91,  the  cyhnder  taken  is  part  of  an 
elbow  with  a  90  degree  turn.    With  the  bow  dividers  set  off  on 


112 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


the  base  the  distances  1-2,  2-3,  3-4,  etc.,  equal  and  small  enough 
so  that  the  chord  of  the  arc  is  equal,  practically,  to  the  arc  itself, 

and  through  these  points 
draw  elements  of  the  cylin- 
der. The  last  step,  11-12, 
may  not  be  equal  to  the 
others  but  this  does  not 
matter.  Now  draw  a  line 
i-i  (it  will  be  convenient 
to  draw  this  line  with  the 
T  square  a  continuation 
of  the  projection  of  the 
base  of  the  cylinder)  and 
divide  it  as  shown  so  that 
the  divisions  are  equal  to 
the  distances  laid  off  on 
the  base  of  the  cylinder. 
Through  these  points  erect  perpendiculars  equal  in  length  to 
the  elements.  Through  the  points  a,  b,  c,  d,  etc.,  thus  found 
draw   a   curve.     The   area   bounded    by   the    points    a-i-i-a 


Fig.  90. 


5    4    3    2    1 


Fig.  91. 

is  the  developed  cylinder.  It  will  be  obvious  from  the  figure 
that  since  the  development  is  symmetrical  on  each  side  of  the 
center  line  1 2-I  only  one-half  of  the  base  need  be  divided  in  order 
to  develop  the  whole  cylinder. 


CYLINDRICAL   SURFACES  II3 

PROBLEMS  ON  CYLINDERS 

187.  The  axis  of  a  cylinder  pierces  H  at  a  point  i^"  in  front  of  V.  A 
second  point  on  this  line,  4''  from  the  first  point,  is  3''  in  front  of  V  and  2^" 
above  H.  A  right  section  of  this  cyhnder  is  a  circle  2"  in  diameter.  Draw 
a  plan  and  elevation  of  this  cylinder  and  show  the  true  shape  and  size  of  its 
base  in  H. 

188.  The  point  0  is  o";  -  2";  o".  The  point  P  is  V;  -4";  -3".  The 
line  OP  is  the  axis  of  a  cyhnder  whose  right  section  is  a  circle  2"  in  diameter. 
Draw  a  plan  and  elevation  of  this  cylinder  and  find  its  base  in  H. 

189.  The  axis  of  a  cylinder  pierces  H  at  a  point  2"  back  of  V  and  pierces 
V  at  a  point  3"  below  H.  Between  these  two  piercing  points  the  axis  is  6" 
long.  Draw  a  plan  and  elevation  of  the  cylinder  showing  its  base  in  H  as  a 
circle  i|"  in  diameter  and  find  its  base  in  V. 

190.  The  axis  of  a  cylinder  pierces  V  4"  below  H,  and  inclines  30  degrees 
to  V  and  45  degrees  to  H.  Draw  a  plan  and  elevation  of  the  cylinder  and 
find  its  base  in  H  when  its  base  in  V  is  a  circle  2"  in  diameter. 

191.  A  3"  circular  air  pipe  crosses  the  corner  of  a  room.  Its  axis  cuts 
the  side  wall  5"  from  the  comer  and  the  front  wall  6"  from  the  comer;  both 
of  these  points  are  8''  below  the  ceiling.  Show  the  tme  size  of  the  holes 
which  will  have  to  be  cut  for  the  pipe. 

192.  The  axis  of  a  right  circular  cyhnder  3''  in  diameter  pierces  H  2" 
back  of  V  and  is  perpendicular  to  H.  Find  the  traces  of  a  plane  which  will 
show  the  angle  at  which  this  cylinder  has  to  be  cut  in  order  to  get  an  ellipse 
whose  axes  are  4"  and  3". 

193.  The  point  O  is  o";  -o";  -if".  The  point  P  is  o";  -4";  if. 
This  line  is  the  axis  of  a  right  circular  cylinder  whose  diameter  is  2^".  The 
cylinder  is  cut  by  a  plane  located  at  4";  135;  240.  Show  a  plan  and  eleva- 
tion of  the  curve  of  intersection  and  its  true  size. 

194.  The  point  O  is  o";  -2";  -o";  the  point  P  is  o";  -2";  -4". 
The  plane  T  is  —4";  +30;  —45.  Find  the  true  shape  and  size  of  the  curve 
cut  from  the  cylinder  whose  axis  is  OP  and  whose  diameter  is  2". 

195.  Draw  the  plan  and  elevation  of  a  cyhnder  whose  base  is  a  2"  circle 
in  H  with  its  center  2"  below  the  G.  L.,  and  whose  axis  is  inclined  so  that  its 
plan  makes  45  degrees  with  the  G.  L.  and  its  elevation  makes  60  degrees 
with  the  G.  L.  Cut  this  cylinder  by  a  plane  which  makes  an  angle  of  75 
degrees  with  the  axis.  Find  the  plan  and  elevation  of  the  curve  of  inter- 
section and  its  true  size. 

196.  The  point  0  is  o'';  -  3";  o'';  the  pomt  Q  is  3";  -5";  -4".  The 
line  OQ  is  the  axis  of  a  cylinder  whose  base  in  H  is  a  circle  2^"  in  diameter. 
The  plane  T  is  4";  —45;  —45.  Draw  a  plan  and  elevation  of  the  inter- 
section of  the  plane  and  cylinder  and  find  its  tme  size. 

197.  The  plane  S  is  o;  —45;  —60.  The  point  O  hes  in  this  plane  at 
—  i";  —2";  x",  and  is  the  center  of  a  2"  circle  whose  plane  is  S.    Draw  the 


114  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

plan  and  elevation  of  the  cylinder  of  which  this  circle  is  a  right  section  and 
find  one  of  its  bases  in  H  or  V. 

198.  The  point  0  is  -2'';  -f]  o".  The  point  Q  is  2";  o";  -f.  The 
line  joining  OQ  is  the  axis  of  a  circular  cyUnder  whose  diameter  is  2".  Draw 
the  plan  and  elevation  of  the  cylinder. 

199.  The  axis  of  a  cylinder  whose  right  section  is  a  2"  circle  pierces  H  3'' 
back  of  V  and  pierces  V  4"  below  H.  Between  these  two  points  the  axis  is 
7"  long.    Draw  a  plan  and  elevation  of  the  cyHnder  and  find  its  base  in  P. 

200.  A  right  circular  cyHnder  whose  diameter  is  2^''  is  cut  at  an  angle 
of  60  degrees  to  its  axis.  Make  a  pattern  for  this  cylinder  assuming  the 
longest  element  of  the  cylinder  to  be  4''. 

201.  A  right  elliptical  cylinder  is  cut  off  by  a  plane  incHning  to  the  axis 
at  such  an  angle  that  the  section  cut  is  a  circle.  The  right  section  of  the 
cyHnder  is  an  elHpse  2'^  by  3''  and  its  longest  element  is  5''.  Lay  out  a 
pattern  for  this  cylinder. 

202.  Make  a  development  of  the  air  pipe  in  Problem  191  and  lay  out  on 
the  development  the  line  of  intersection  between  the  pipe  and  the  walls. 

203.  Make  the  patterns  for  a  four  piece  elbow  of  6"  pipe.  The  elbow 
turns  an  angle  of  90  degrees,  with  a  radius  of  center  line  12"  long. 

204.  The  axis  of  a  3"  circular  cylinder  pierces  H  4''  back  of  V  and  pierces 
V  3"  below  H;  between  these  two  piercing  points  the  axis  is  5'' long.  Make 
a  pattern  of  the  cyHnder  between  its  H  and  V  bases. 

205.  A  Hne  of  8''  pipe  slopes  at  the  rate  of  4'  in  every  10'  and  runs  due 
northeast.  At  a  point  o";  —6";  o''  the  pipe  meets  the  horizontal  floor  it  is 
to  drain.  Make  a  pattern  showing  how  the  last  section  of  pipe  must  be  cut 
to  fit  into  the  floor. 

206.  Make  the  patterns  for  a  five  piece  elbow  of  8"  pipe.  The  pipe 
turns  an  angle  of  105  degrees,  with  a  radius  of  16". 


CHAPTER  XIV 
CONICAL   SURFACES 

97.  A  conical  surface  is  represented  by  the  projections  of  some 
curve  of  its  surface  and  the  limiting  elements  which  meet  at  the 
apex.  As  in  the  case  of  cylinders  this  curve  is  usually  taken  in 
H  or  V  and  is  the  locus  of  the  piercing  points  of  the  elements. 

98.  A  cone  is  represented  by  the  projections  of  any  one  of  its 
sections  and  the  limiting  elements.  For  convenience  the  section 
taken  is  usually  the  base  in  H  or  V,  and  the  projections  of  the 
limiting  elements  are  found  by  joining  the  limiting  points  in  the 
base  with  the  projections  of  the  apex.  Thus,  in  Fig.  92,  the  base 
is  an  ellipse  in  H  whose  center  is  at  0,  and  the  apex  of  the  cone 
is  given  at  Q.  To  find  the  projections  of  the  limiting  elements 
for  the  plan  views  of  the  cone  tangents  are  drawn  from  q  to  the 
ellipse;  and  to  find  the  projections  of  the  limiting  elements  for 
the  elevation  of  the  cone  lines  are  drawn  from  q'  to  the  extreme 
points  of  the  elevation  of  the  base.  It  will  be  obvious  that  the 
limiting  elements  of  the  elevation  are  not  projections  of  the 
limiting  elements  of  the  plan  view. 

99.  Cones  are  identified  by  their  bases.  Thus:  a  cone  with 
a  circular  base  or  a  cone  with  an  elliptical  base,  etc.  As  in  the 
case  of  cylinders  a  cone  with  a  circular  base  will  be  elliptical  in 
section  when  its  axis  is  inclined  to  the  plane  of  the  base,  and  a 
cone  with  an  elliptical  base  may  be  either  elliptical  or  circular 
in  section  according  to  the  angle  of  inclination. 

A  right  cone  is  one  whose  axis  is  perpendicular  to  the  plane 
of  the  base.  Right  cones  may  be  either  circular  or  elliptical  in 
section  and  are  named  accordingly,  as :  a  right  circular  cone,  or 
a  right  elliptical  cone. 

100.  To  assume  a  point  on  the  surface  of  the  cone  either 
projection  of  a  point  may  be  assumed  on  the  corresponding  pro- 

"S 


ii6 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


jection  of  some  element  and  the  other  projection  of  the  point 
found  by  locating  it  on  the  other  projection  of  the  same  element. 
Thus,  in  Fig.  92,  the  projection  y'  is  assumed  on  the  projection 
of  the  element  q'r'  and  the  projection  y  then  will  be  found  on  qr. 
It  will  be  noted,  as  in  the  case  of  cylinders,  that  there  are  two 
elements  of  the  cone  whose  elevations  coincide  with  this  line,  one 
being  QR  and  the  other  QP,  so  that  it  is  possible  to  have  two 
points  on  the  cone  whose  elevations  coincide  at  x',  one  being  Y 


Fig.  92. 


Fig.  93. 


and  the  other  X.  In  given  problems  the  data  are  usually  given 
which  determine  which  point  is  intended,  otherwise  two  solutions 
to  such  problems  are  possible. 

In  Fig.  93  a  similar  problem  is  shown.  In  this  case  the  plan 
view  of  the  point  is  assumed  and  its  elevation  located  at  one  of 
the  points  x'  or  y',  depending  on  which  side  of  the  cone  the  point 
is  to  be  taken. 

loi.  Proposition  26.  Given  the  axis  of  a  cone,  the  size  and 
location  of  its  right  section  to  find  the  plan  and  elevation  of  the 
cone. 

Discussion.  If  the  cone  be  revolved  parallel  to  a  plane  of 
projection,  the  given  right  section  may  be  drawn  in  its  given 
location.    If,  then,  elements  be  drawn  through  this  section  from 


CONICAL   SURFACES 


117 


the  apex  the  base  of  the  cone  in  H  or  V  may  be  found  by  finding 
where  the  elements  pierce  H  or  V. 

Construction.  In  Fig.  94,  OQ  is  the  given  axis,  and  P  the 
given  point  on  this  axis  where  it  pierces  the  given  right  section. 
The  right  section  in  this  construction  is  taken  as  a  circle  whose 
diameter  is  equal  to  x"y''.     Revolve  OQ  parallel  to  V;  in  this 

r5 


9'  r 


Fig.  94. 


position  it  is  projected  at  oq^  and  o'q"  and  P  falls  at  p^  and  p". 
Draw  x"y"  equal  to  the  given  diameter  of  the  right  section  per- 
pendicular to  o'q" ;  this  line  will  be  the  elevation  of  the  right  sec- 
tion, since  OQ  is  parallel  to  V,  and  the  ellipse  x^y^  will  be  the  plan 
view.  Now  through  x''  and  y''  draw  the  elements  projected 
on  V  at  q"d"  and  q"a",  and  on  H  at  q^d^  and  q,a^.  These 
elements  pierce  H  at  a^  and  d^,  which  points  will  be  the  ends  of 
the  major  axis  of  the  base  in  H.     To  find  the  minor  axis  of  the 


Il8  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

base  bisect  a^d^  by  the  line  whose  direction  is  b^c^.  The  length 
of  the  Hne  is  as  yet  unknown  but  whatever  this  may  be,  since  it 
is  perpendicular  to  V,  the  line  will  be  projected  on  V  at  the  point 
b'V.  Join  b''c"  with  q";  this  will  be  the  elevation  of  the  two 
elements  which  pierce  H  at  the  ends  of  the  minor  axis.  Now 
these  elements  cut  the  curve  of  right  section  in  its  revolved 
position  at  M  and  N,  and  if  through  these  points  elements  be 
drawn  they  will  pierce  H  at  c^  and  d^,  or  the  ends  of  the  minor 
axis.  The  axis  may  now  be  revolved  to  its  original  position,  and 
the  true  position  of  these  axes  located.  The  curve  in  the  base 
may  be  constructed  and  the  Hmiting  elements  drawn. 

102.  Proposition  27.  To  find  the  curve  cut  from  any  given 
cone  by  any  given  plane. 

Discussion.  Through  the  apex  of  the  cone  pass  auxiliary 
planes  intersecting  the  cone  and  the  given  plane.  These  planes 
cut  elements  from  the  cone  and  Unes  from  the  given  plane. 
Since  these  elements  and  lines  lie  in  common  planes  they  will 
intersect  in  points  which  will  be  common  to  both  the  cone  and 
the  given  plane.  If  a  sufficient  number  of  auxiliary  planes  be 
passed  enough  points  may  be  found  to  plot  the  curve  of  inter- 
section. 

Construction.  When  the  axis  of  the  cone  is  perpendicular  to 
H  or  V  and  its  base  is  a  circle.  In  Fig.  95,  sSs'  is  the  given  plane 
cutting  the  right  circular  cone  whose  apex  is  at  Q.  Through  the 
apex  Q  pass  the  plane  R  perpendicular  to  Ss.  This  plane  cuts 
from  the  plane  S  the  line  AB,  which  Hne  pierces  the  surface  of  the 
cone  at  A  and  B.  Since  this  hne  is  the  longest  line  which  can  be 
drawn  common  to  both  the  cone  and  the  plane  S  it  will  be  the 
major  axis  of  the  curve  of  intersection.  Now  through  P,  the 
middle  point  of  AB,  draw  CO  in  plane  S  and  perpendicular  to 
AB.  This  Hne  CO  pierces  the  surface  of  the  cone  at  the  points 
C  and  D.  Since  it  is  the  shortest  Hne  which  can  be  drawn  com- 
mon to  both  cone  and  plane  S,  and  since  it  is  perpendicular  to 
AB,  CD  will  be  the  minor  axis  of  the  curve  of  intersection.  To 
find  its  true  size  revolve  the  axes  of  the  curve  into  V  about  s'S 
and  construct  the  ellipse  AvByCvDy. 

If  the  projections  of  this  curve  of  intersection  are  required  they 


CONICAL  SURFACES 


119 


may  be  found  by  passing  other  auxiliary  planes,  such  as  R, 
through  the  apex  of  the  cone  and  finding  points  by  the  method 
used  to  find  points  A  and  B. 


Fig.  95. 

Construction.  When  the  axis  of  the  cone  is  oblique  to  H  and 
V,  the  base  being  any  plane  curve. 

In  Fig.  96,  S  is  the  given  plane  cutting  the  cone  whose  apex 
is  Q.     Pass  auxiliary  planes  R,  U,  V,  etc.,  through  O  perpendicu-. 


I20 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


lar  to  H.  The  auxiliary  plane  R  cuts  from  the  cone  one  element 
and  from  the  plane  S  the  line  OM.  This  line  OM  intersects  this 
element  at  F,  one  point  on  the  curve  of  intersection.  In  like 
manner  the  plane  U  determines  the  point  E  of  the  curve.     Now 


Fig.  96. 

in  problems  of  this  character  it  generally  happens  that  the 
auxiliary  planes  run  off  the  paper,  as,  for  example,  V,  making  it 
impossible  to  find  points  common  to  the  V  traces,  as  m'  and  n'. 
To  avoid  the  difficulty  which  arises  the  following  method  of 
finding  the  intersections  of  S  with  the  cutting  plane  is  convenient. 


CONICAL  SURFACES  121 

Since  the  auxiliary  planes  all  pass  through  the  apex  of  the 
cone  and  are  perpendicular  to  H,  they  will  have  a  line  in  com- 
mon which  passes  through  Q  perpendicular  to  H.  Now  this 
line  pierces  the  given  plane  S  in  a  point  which  is  common  to  S 
and  all  the  auxiliary  planes.  This  point,  therefore,  is  common 
to  all  the  lines  cut  from  S  by  the  auxiliary  planes.  Through  this 
point  and  the  intersections  of  the  H  traces  hues  of  intersection 
between  the  planes  may  be  drawn  without  using  the  V  traces. 
This  point  in  the  given  case  is  O.  To  find  it  assume  a  point  O 
in  the  plane  S,  having  its  plan  view  at  q  (Article  26).  Since  the 
line  common  to  all  the  auxiliary  planes  is  perpendicular  to  H 
through  the  apex  its  plan  view  will  coincide  with  q. 

Now  the  auxiliary  plane  V  cuts  from  the  plane  S  the  line  OK 
and  from  the  cone  two  elements;  this  line  and  the  elements 
intersect  at  B  and  D  on  the  required  curve.  In  like  manner 
other  points  may  be  determined  and  the  curve  of  intersection 
drawn.  To  find  the  true  shape  and  size  of  the.curve  it  is  neces- 
sary to  revolve  it  into  H  or  V  about  the  corresponding  trace  and 
to  locate  a  number  of  points,  as  AyByCv,  etc. 

103.  Proposition  28.  Given  any  cone  to  find  the  shape  and 
size  of  its  right  section  at  any  given  point. 

Discussion.  If  the  cone  be  revolved  parallel  to  H  or  V  the 
right  section  may  be  drawn  through  the  given  point.  In  this 
position  the  axes  of  the  right  section  will  be  projected  on  H  or 
V  in  their  true  size. 

Construction.  In  Fig.  97,  the  given  cone  has  a  circular  base 
in  V  and  an  apex  at  0.  About  the  center  of  the  base  revolve 
the  cone  parallel  to  H,  the  projections  of  the  apex  falling  at  o, 
and  o''.  Now  find  OQ,  the  axis  of  the  cone,  by  bisecting  the 
angle  of  the  apex  and  on  this  line  locate  the  given  point  P,  at 
which  the  section  is  to  be  taken.  Through  P,  pass  a  plane  R 
perpendicular  to  the  revolved  position  of  OQ.  This  plane  con- 
tains the  revolved  position  of  the  right  section,  and  the  points 
C  and  D,  where  the  limiting  elements  pierce  R,  are  the  ends 
of  the  minor  axis.  Since  CD  is  parallel  to  V  and  the  axes  are 
perpendicular,  the  major  axis  will  be  projected  on  H  as  a  point 
at  p^  and  on  V  as  a  line  through  p"  perpendicular  to  c''d". 


122 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


Now  through  p^  draw  the  projection  of  the  elements  which  pass 
through  the  ends  of  the  major  axis;  both  of  these  elements  are 
projected  in  H  at  o.q^  and  pierce  V  at  x"  and  y" .  o"x"  and 
o"y"  are  the  elevations  of  the  elements  which  pass  through  the 
ends  of  the  major  axis  at  a"  and  b".     a"b''  and  c''d''  are  ele- 


FiG.  97. 


vations,  then,  of  the  axes  of  the  right  section  at  P,  and  by 
revolving  them  into  V  about  r'R  the  true  size  of  the  curve  may 
be  drawn  at  AyByCvDv. 

104.  If  the  base  of  the  cone  is  an  elUpse  whose  major  axis 
coincides  in  direction  with  the  projection  of  the  axis  of  the  cone, 
this  same  method  may  be  used  to  find  the  right  section.  In  such 
a  case,  however,  it  is  possible  for  the  section  to  be  a  circle  and  if 
the  cone  inclines  so  that  the  plane  R  cuts  a  Hne  c^d^  from  the 
cone  equal  to  the  line  a"b"  such  will  be  the  case. 

When  this  method  will  not  apply  the  right  section  may  be 


CONICAL   SURFACES 


123 


found  as  in  Article  102  by  taking  the  cutting  plane  perpen- 
dicular to  the  axis  of  the  cone. 

105.  As  in  the  case  of  cylinders  the  fact  that  an  oblique  plane 
cuts  a  symmetrical  curve  from  a  right  cone  may  be  utilized  in 


Fig.  98. 


Fig.  99. 


constructing  conical  elbows.     Fig.  98  shows  a  typical  case  when 
the  top  of  a  given  cone  is  cut  off,  turned  through  180  degrees, 


and  set  back.  The  axes  of  the  two  sections  not  only  turn  an 
angle  but  they  no  longer  meet.  The  reason  for  the  latter  point 
will  be  clear  from  an  inspection  of  Fig.  99. 


124 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


Fig.  loi  shows  an  elbow  made  up  of  sections  of  the  same 
cone,  the  cone  being  divided  as  shown  in  Fig.  loo.  Since  the 
axis  of  the  cone  does  not  pierce  the  plane  of  each  joint  in  the 
center  of  the  plane  of  the  joint  the  axes  will  have  to  be  offset  as 
shown. 

A  simple  method  of  drawing  such  elbows  is  shown  in  Fig.  103, 

which  is  a  design  for  a  ship's  ventilator  of  three  conical  and  one 

cylindrical  sections,  whose  axes  are  given  by  the  broken  line 

M 


Fig.  102. 

ABCDE.  The  problem  is  to  find  the  size  of  the  right  cones 
which  will  fit  together  with  plane  joints. 

The  principle  by  which  the  problem  may  be  worked  may  be 
stated  as: 

106.  Proposition  29.  To  find  the  size  of  a  cone  or  cylinder 
which  will  fit  another  cone  or  cylinder  with  a  plane  joint,  the 
axes  and  the  size  of  one  cone  or  cylinder  being  given. 

Discussion.  Since  an  oblique  plane  cuts  an  ellipse  from  both 
a  right  cone  and  a  right  cylinder,  the  problem  is  to  find  where  to 
place  this  plane  so  that  the  sections  cut  from  the  surface  will  be 
equal.  In  Fig.  102  the  right  cylinder  whose  axis  is  AO  is  to  form 
an  elbow  with  a  right  cone  whose  axis  is  BO,  and  whose  right 
section  at  B  is  the  circle  shown.  Now  if  at  O,  the  intersection  of 
the  axes,  a  sphere  be  drawn  to  which  the  given  cylinder  is  tan- 
gent, and  a  cone  be  drawn  with  BO  as  an  axis  tangent  to  the 
sphere,  this  cone  will  join  the  cylinder  in  an  ellipse  whose  plane 


CONICAL  SURFACES 


125 


is  shown  as  the  joint  in  the  elbow,  MN.  Since  the  cone  and  the 
cylinder  are  each  tangent  to  a  common  sphere  they  will  have  one 
curve  of  intersection  in  common. 

Construction.  To  apply  this  principle  to  the  drawing  of  the 
ventilator  cowl  made  up  of  conical  sections,  spheres  must  be 
drawn  at  B,  C,  and  D  as  shown  in  Fig.  103.  The  diameter  of 
the  sphere  at  B  must  be  equal  to  the  diameter  of  the  cylinder; 
the  diameter  of  the  spheres  at  C  and  D  are  to  be  assumed  so  that 


Fig.  103. 


the  increase  in  size  in  the  sections  of  the  cowl  is  fairly  uniform. 
Tangent  to  these  spheres  are  drawn  the  limiting  elements  of  the 
cones  and  where  these  pairs  of  limiting  elements  intersect  will  be 
the  joints.  The  opening  at  E  is  usually  given  so  that  the  hmiting 
elements  of  the  last  section  are  drawn  from  the  ends  of  the  given 
diameter  of  the  opening  at  E  to  the  sphere  whose  center  is  at  D. 

107.   Proposition  30.     To  develop  any  given  cone. 

Discussion.  Draw  lines  from  a  common  point  —  the  apex  — 
equal  in  length  to  the  elements  of  the  cone,  and  spaced  according 
to  their  distances  apart  on  some  curve  of  the  surface. 


126 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


Construction.  When  the  given  cone  is  a  right  circular  cone. 
In  Fig.  104  the  base  of  the  given  cone  is  a  circle  in  H,  the  apex  is 
at  O,  and  the  cone  has  been  cut  off  along  the  plane  of  a'g'.  Since 
the  cone  is  right  all  the  elements  from  the  apex  to  the  base  are 


Fig.  104. 

equal  in  length;   therefore,  with  o'  as  a  center  and  o'l'  as  a 

radius  describe  the  arc  i 16,  etc.,  and  at  any  convenient 

point  draw  the  Une  o'gi  equal  to  o'g'i'.  This  Hne  will  be  the 
first  element  in  the  development.  Now  divide  the  circle  of  the 
base  into  a  convenient  number  of  parts  and  step  these  off  on 


CONICAL  SURFACES 


127 


the  developed  base  as  i,  2,  3,  4,  5,  6,  etc.  Find  where  each  of  the 
elements  pierces  the  plane  through  dJg'  and  find  the  true  distance 
from  0  to  each  point.     Lay  this  distance  off  on  the  development 


—7^ 
// 

"        '     /     /  // 
II         I      I 

II      I    1    I  H 

w  \  \  \  \ ''? 

\\  \  \  \  \- 


Fig.  105. 


as  at  g,  f,  e,  d,  etc.  In  the  figure  only  a  few  points  have  been 
worked  out  to  avoid  the  confusion  of  lines  but  in  practice  the 
elements  should  be  spaced  about  as  shown  by  the  points  i,  2, 
3j  4,  etc. 


128  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

Since  the  development  will  be  symmetrical  about  a  center  line 
Oi6  the  two  parts  may  be  constructed  simultaneously. 

Construction.  Method  of  Triangulation.  When  the  given 
cone  is  oblique.  In  Fig.  105  the  given  cone  is  taken  with  its 
axis  parallel  to  V  with  its  base  in  H  and  its  apex  at  O.  Lay 
off  o'8  equal  to  element  08.  Now  with  the  true  length  of  the 
arc  7-8  as  a  radius  and  8  as  a  center  strike  an  arc,  and  with  o' 
as  a  center  and  the  true  length  of  0  7  as  a  radius  strike  a  second 
arc  intersecting  the  first  at  7.  0^7  then  is  the  position  in  the 
development  of  the  element  O7.  In  practice  the  distance  8-7 
must  be  made  small  enough  so  that  the  triangle  0^87  will  He 
on  the  surface  of  the  cone;  in  the  figure  these  lines  are  taken 
far  apart  to  avoid  confusion. 

In  like  fashion  with  0^7  as  one  side  construct  triangle  0^76 
equal  to  the  triangle  O76  on  the  cone,  and  proceeding  this  way 
the  entire  surface  of  the  cone  may  be  developed  by  dividing  it  up 
into  small  triangles  and  laying  these  out  in  their  true  size  and 
relation. 

This  method  is  the  method  by  which  most  developments  are 
made.  It  will  apply  to  the  development  of  any  object  whose 
surface  is  capable  of  being  divided  into  triangles. 


PROBLEMS  ON  CONES 

207.  The  axis  of  a  cone  is  3^"  long.  It  pierces  H  if"  in  front  of  V  and 
inclines  45  degrees  to  H.  The  apex  of  the  cone  is  3"  in  front  of  V.  A  right 
section  of  the  cone  2"  from  the  apex  is  a  circle  \"  in  diameter.  Draw  a  plan 
and  elevation  of  the  cone  and  find  its  base  in  H. 

208.  The  point  0  is  o";  -3'';  o".  The  point  Q  is  -3'';  -5'';  -4". 
The  line  OQ  is  the  axis  of  a  cone  whose  right  section  4"  from  the  apex  Q  is  a 
circle  i\"  in  diameter.  Draw  the  plan  and  elevation  of  the  cone  and  show 
its  base  in  H. 

209.  The  axis  of  a  cone  pierces  H  3"  in  front  of  V.  The  apex  of  the 
cone  is  3"  above  H  and  2"  behind  V  and  the  axis  from  the  apex  to  the  H 
piercing  point  is  5"  long.  The  base  of  the  cone  in  H  is  a  circle  2\"  in 
diameter.     Find  the  base  of  the  cone  in  V. 

210.  The  point  0  is  -f;  -3'';  o" .  The  point  Q  is  2";  -3'';  -4". 
The  H  base  of  this  cone  whose  axis  is  OQ  is  a  circle  2\"  in  diameter.  Find 
the  base  of  the  cone  in  P. 


CONICAL   SURFACES  1 29 

211.  The  point  0  is  -2";  o";  -3".  The  point  Q  is  2";  -4";  -f. 
The  base  of  the  cone  whose  axis  is  OQ  is  a  1"  circle  in  P.  Find  the  base  of 
this  same  cone  in  V, 

212.  A  conical  lamp  shade  whose  height  is  8"  is  hung  from  the  center 
of  a  g'  by  12'  room  with  its  apex  7'  6"  above  the  floor.  The  base  of  the 
shade  is  1 2".  Find  the  lighted  areas  on  the  walls  and  floor  of  the  room  made 
by  the  cone  of  light  reflected  from  the  shade. 

213.  The  axis  of  a  right  cone  is  perpendicular  to  H  2\"  behind  V;  the 
base  of  this  cone  in  H  is  a  3"  circle  and  its  altitude  is  4".  Cut  this  cone  by 
three  planes:  one  at  an  angle  to  its  axis  cutting  all  of  the  elements;  one 
parallel  to  the  axis;  and  one  parallel  to  one  element  of  the  cone.  Show  the 
true  shape  and  size  of  each  of  these  curves  of  intersection  and  investigate 
its  properties. 

214.  Draw  a  plan  and  elevation  of  the  cone  in  Problem  213  and  cut  the 
cone  by  any  plane  oblique  to  both  H  and  V.  Show  the  true  shape  and  size 
of  the  curve  of  intersection. 

215.  The  point  O  is  o";  -o";  -2\".  The  point  Q  is  o";  -2I")  -2\'\ 
The  base  of  the  cone  whose  axis  is  OQ  is  a  circle  2\"  in  diameter  in  V.  The 
cone  is  cut  by  plane  T  =  3^";  —30;  —60.  Find  the  true  shape  and  size  of 
the  curve  of  intersection  and  its  plan  and  elevation. 

216.  The  point  0  is  3I";  -2\")  o".  The  point  Q  is  if";  -4!";  -3!". 
The  base  of  this  cone  is  a  circle  3"  in  diameter  in  H.  The  cone  is  cut  by  a 
plane  which  is  o";  +30;  —60.  Find  the  true  shape  and  size  of  the  curve 
of  intersection. 

217.  The  point  O  is  3 r;  -2h"',  -o".  The  point  Q  is  o";  -2^";  -2^". 
The  base  in  H  of  this  cone  is  a  circle  2^"  in  diameter.  The  plane  T  is 
o";  —60;  —45  and  cuts  the  cone.  Find  the  true  shape  and  size  of  the  curve 
of  intersection. 

218.  Draw  the  plan,  elevation,  and  left  end  view  of  a  right  cone  whose 
base  is  a  i^"  circle  lying  in  plane  T  —  5I";  60;  30;  with  its  center  at  O 
~3l"')  if";  x"  and  whose  altitude  is  3". 

219.  A  cone  of  revolution  is  located  in  the  third  quadrant  with  its  base 
in  plane  T  —  2§";  —150;  —120  with  its  center  at  the  point  O  —5'';  x"; 
—  1 1".  The  base  is  a  circle  2"  in  diameter  and  the  altitude  is  3^".  Draw 
three  views  of  the  cone. 

220.  Lay  out  the  patterns  for  a  conical  hopper.  Diameter  of  large 
opening  12";  diameter  of  small  opening  4'';  distance  between  openings  6". 

221.  A  20"  smoke  stack  passes  through  a  roof  which  inchnes  30  degrees 
with  the  horizontal.  A  conical  canopy  forms  the  flashing  between  the  roof 
and  stack.  Make  a  pattern  of  this  canopy  with  the  shortest  element  6" 
long. 

222.  Lay  out  a  pattern  for  the  cone  in  Problem  216  between  the  base 
and  the  intersection  with  the  plane. 


130  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

223.  The  point  0  is  o")  3";  o''.  The  point  Q  is  o";  -2;  -4''.  The 
line  OQ  is  the  axis  of  a  cone  whose  base  in  H  is  a  2>"  circle.  Make  a  pattern 
for  that  portion  of  the  cone  lying  between  H  and  V. 

224.  Make  a  pattern  for  a  sink  strainer.  This  is  in  the  form  of  one 
quarter  of  a  cone  whose  opening  is  a  16''  circle  and  whose  height  is  8". 

225.  Make  the  patterns  for  a  conical  elbow  of  three  sections  to  turn  an 
angle  of  90  degrees.  Diameter  of  large  opening  10'';  diameter  of  small 
opening  4";  radius  of  center  line  12''. 

226.  Design  a  conical  elbow  of  four  sections  for  a  120  degree  turn  to 
connect  a  pipe  of  12"  diameter  with  another  pipe  of  d"  diameter,  radius  of 
center  line  20". 


CHAPTER  XV 
INTERSECTION    OF   SURFACES 

1 08.  Proposition  31.  Given  two  plane  surfaces  to  find  their 
lines  of  intersection. 

Discussion:  First  Method.  Find  where  the  edges  of  one 
surface  pierce  the  planes  of  the  faces  of  the  other  surface.  These 
points  when  joined  in  proper  order  determine  the  line  of  inter- 
section. 

Construction.  In  Fig.  106  the  right  hexagonal  prism  is 
intersected  by  a  rectangular  prism  whose  edges  are  oblique  to  H 
and  V.  Find  where  these  edges  pierce  the  faces  of  the  hexagonal 
pyramid  and  join  the  points  in  the  proper  order.  In  the  given 
case  the  rectangular  prism  pierces  the  hexagonal  prism,  thus 
giving  two  intersections:  the  entering  one  at  ABCD  and  the  de- 
parting one  at  EFGH.  The  prisms  have  been  assumed  as  solids 
in  the  drawing. 

Discussion:  Second  Method.  Find  the  intersections  of  the 
planes  of  the  faces.  These  lines  will  be  the  edges  of  the  inter- 
section of  the  two  surfaces. 

Construction.  In  Fig.  107  a  rectangular  hopper,  base  in 
H  1-2-3-4  and  apex  O,  intersects  a  rectangular  pipe.  The  face 
of  the  hopper,  1-4-O,  lies  in  the  planer  Rr';  and  this  plane 
intersects  U  —  the  plane  of  the  upper  face  of  the  pipe  —  in  the 
Une  PQ.  The  portion  of  PQ  between  the  edges  O4  and  Oi  — or 
AD  —  is  one  side  of  the  opening  in  the  pipe  made  by  the  hopper. 
In  like  manner  CB  may  be  obtained  by  finding  the  intersection 
of  plane  U  with  plane  vVv';  and  with  these  four  points.  A,  B,  C, 
D,  the  intersection  may  be  drawn. 

109.  Proposition  32.  To  find  the  intersection  of  two  given 
cylinders. 

Discussion.  Pass  auxiliary  planes  through  the  cyHnders  cut- 
ting elements  from  each.     These  elements,  being  in  a  common 

131 


132 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


plane,  intersect  in  points  common  to  both  cylinders,  or  points 
on  the  required  curve  of  intersection. 


Fig.  io6. 

Construction:  First  Case.  When  the  bases  of  the  given 
cylinders  lie  in  the  same  plane. 

In  Fig.  io8  one  cylinder  is  perpendicular  to  H  and  the  other 
oblique  to  H  and  V;  both  have  circular  bases  in  H.    Pass  plane 


INTERSECTION  OF  SURFACES 


133 


Fig,  107. 


134 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


R  through  both  cylinders  cutting  elements  from  each.  These 
elements  intersect  at  A,  B,  C,  and  D,  as  shown.  If  enough  auxil- 
iary planes  similar  to  R  are  passed  a  sufficient  number  of  points 
may  be  obtained  to  draw  a  curve  of  intersection. 


Fig.  io8. 

Construction:  Second  Case.  When  the  bases  of  the  given 
cylinders  are  in  different  planes. 

In  Fig.  109  the  larger  cylinder  has  a  circular  base  in  H  and  is 
parallel  to  V;  the  smaller  has  a  circular  base  in  V  and  is  parallel 
to  H.  Pass  auxiliary  planes,  as  S,  through  the  cylinders  parallel 
to  the  axis  of  each  and  cutting  elements  from  each.     These  ele- 


INTERSECTION  OF  SURFACES 


135 


ments  intersect  in  points  of  the  curve  of  intersection,  as  D  and 
B.  To  find  the  direction  of  the  auxiliary  planes  pass  any  plane 
through  the  axis  of  one  cylinder  parallel  to  the  axis  of  the  other. 
If  the  auxiHary  planes  be  drawn  parallel  to  this  plane  they  will 


109. 


cut  elements  from  the  cylinders  since  they  are  parallel  to  the 
axis  of  each  cyHnder. 

no.   Something  of  the  nature  of  the  curve  of  intersection  may 
be  determined  in  advance.    Thus,  in  Fig.  109,  plane  R  is  ob- 


136  ESSENTIALS   OF  DESCRIPTIVE  GEOMETRY 

viously  the  first  auxiliary  plane  which  can  cut  elements  from  the 
smaller  cylinder,  and  the  plane  T  is  obviously  the  last  auxiliary 
plane  which  can  cut  the  smaller  cylinder.  Therefore,  since  all 
of  the  elements  of  the  smaller  cylinder  pierce  the  surface  of  the 
larger  one,  it  becomes  apparent  that  the  smaller  one  completely 
enters  the  larger.  On  the  other  hand,  if  the  plane  rRr'  failed  to 
cut  elements  from  the  larger  cylinder,  then  it  would  be  apparent 
that  all  of  the  elements  of  the  smaller  cylinder  did  not  pierce  the 
larger,  and  the  resulting  curve  would  be  more  Hke  the  one  in 
Fig.  108.  By  the  use  of  limiting  planes  in  this  fashion  the 
character  of  the  required  intersection  may  be  somewhat  prede- 
termined and  the  problem  attacked  with  greater  appreciation. 

111.  The  visible  and  invisible  portions  of  the  curve  of  inter- 
section should  be  shown  by  using  solid  and  dotted  lines.  To 
determine  which  points  are  visible  it  must  be  remembered  that 
surfaces  are  not  transparent,  nor  are  they  solids  —  unless  so 
specified  —  and  that  when  one  looks  into  the  end  of  a  cone  or  a 
cylinder  the  base  is  not  considered  an  obstruction  to  the  view. 
A  line  which  enters  a  surface  is  visible  only  up  to  the  entering 
point,  invisible  within  the  surface,  and  visible  again  upon  leaving 
the  surface.  The  simplest  method  of  locating  visible  points  on 
a  curve  of  intersection  of  two  surfaces  is  to  locate  first  the  visi- 
ble elements  of  the  surface  and  of  course  points  on  these  ele- 
ments are  also  visible.  It  is  not  necessary  to  actually  draw  these 
elements  but  with  a  little  practice  and  study  one  can  readily 
locate  the  visible  and  invisible  points  by  inspection. 

112.  Proposition  33.  Given  two  cones  to  find  their  curve  of 
intersection. 

Discussion.  Pass  auxiliary  planes  through  the  apices  of  the 
two  cones  cutting  elements  from  each.  These  elements  being 
in  the  same  plane  will  intersect  each  other  in  points  common  to 
both  surfaces,  or  points  on  the  curve  of  intersection. 

Construction.  When  the  bases  of  the  given  cones  are  in  the 
same  plane,  as  in  Fig.  no. 

Draw  a  line  through  the  apices  of  the  cones  and  find  where  it 
pierces  H  and  V.  These  two  points  will  be  common  to  all  the 
auxiliary  planes.     In  the  case  given  in  Fig.  no  this  line  is  par- 


INTERSECTION  OF  SURFACES 


137 


Fig.  I 10. 


138 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


allel  to  H  so  that  all  the  H  traces  of  the  auxiliary  planes  will  be 
parallel  to  it.  The  plane  S  cuts  one  element  from  the  right 
hand  cone  and  two  elements  from  the  other;  the  elements  in- 
tersect at  B  and  D,  two  points  on  the  curve  of  intersection.  In 
like  manner  other  points  may  be  located  but  all  the  points 
common  to  both  surfaces  will  lie  in  auxiHary  planes  between 
planes  S  and  T. 


Fig.  III. 


Construction.  When  the  bases  of  the  given  cones  are  in 
different  planes. 

In  Fig.  Ill  two  right  cones,  one  with  its  base  in  H  and  the 
other  with  its  base  in  P,  intersect.  As  before,  the  apices  are 
joined;  this  line  OQ  is  common  to  all  the  auxiliary  planes  and 
its  H  piercing  point,  0,  is  common,  therefore,  to  all  the  H  traces 
and  its  P  piercing  point,  Q,  to  all  the  P  traces.     As  before,  locate 


INTERSECTION  OF  SURFACES 


139 


the  limiting  planes,  S  and  R,  and  find  the  points  common  to  both 
surfaces. 

113.  Proposition  34.  Given  a  cone  and  a  cylinder  to  find 
their  intersection. 

Discussion.  Pass  planes  through  the  apex  of  the  cone  parallel 
to  the  elements  of  the  cylinder.     These  planes  will  cut  elements 


Fig.  112. 


from  each  surface,  and  since  these  elements  are  in  a  common 
plane  they  will  intersect  in  points  common  to  both  surfaces. 

Construction.  When  the  bases  of  the  given  surface  are  in 
the  same  plane. 

In  Fig.  112  the  given  cone  and  cylinder  have  circular  bases  in 
V  and  are  oblique  to  both  H  and  V.     Through  0,  the  apex  of 


I40 


ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 


the  cone,  draw  a  line  OQ  parallel  to  the  axis  of  the  cyhnder. 
This  line  will  be  common  to  all  the  auxiliary  planes. 

The  first  plane  whose  V  trace  passes  through  Q  and  which 
cuts  both  the  cone  and  cylinder  is  R.  Plane  R  cuts  one  element 
from  the  cone  and  two  elements  from  the  cylinder;  these  elements 


Fig.  113. 

intersect  at  D  and  C  which  are  two  points  on  the  required  inter- 
section. The  last  plane  which  cuts  both  surfaces  is  plane  S. 
This  plane  cuts  only  one  element  from  the  cylinder  and  two  from 
the  cone;  these  elements  intersect  at  A  and  B  which  are  two 
more  points  on  the  required  curve.     Between  these  limiting 


INTERSECTION  OF  SURFACES  141 

planes  other  auxiliary  planes  may  be  passed  in  sufficient  number 
to  completely  locate  the  curve. 

Construction.  When  the  bases  of  the  given  surfaces  are  in 
different  planes. 

In  Fig.  113  the  base  of  the  cylinder  is  in  P  and  its  elements  are 
parallel  to  the  G.  L.,  and  the  base  of  the  cone  is  in  H  and  its  axis 
is  perpendicular  to  H.  The  construction  will  be  obvious  from  a 
study  of  the  drawing. 

PROBLEMS  ON  THE  INTERSECTION  OF  PLANE  SURFACES 

227.  The  axis  of  an  hexagonal  prism  whose  sides  are  2"  wide  pierces  V  2" 
below  H,  and  is  perpendicular  to  V.  The  axis  of  a  square  prism  whose 
sides  are  \"  wide  pierces  P  2"  below  H  and  2"  back  of  V,  and  is  parallel 
to  the  G.  L.  The  diagonals  of  the  right  section  of  the  square  prism  are 
perpendicular  and  parallel  to  H.  Draw  a  plan  and  elevation  of  the  two 
prisms  showing  their  intersection. 

228.  The  base  of  a  square  pyramid  is  parallel  to  and  \"  below  H.  The 
base  is  2"  by  3''  with  its  3"  edge  parallel  to  V  i"  back  of  V.  The  apex  of 
the  pyramid  is  2"  back  of  V,  2\"  below  H,  and  in  a  profile  plane  \"  to  the 
right  of  the  right  edge  of  the  base.  The  axis  of  a  i"  by  \"  prism  pierces 
H  i|^''  to  the  right  of  the  profile  plane  of  the  apex,  2"  back  of  V,  inclines 
45  degrees  to  H,  and  is  parallel  to  V.  Draw  a  plan  and  elevation  showing 
the  intersection. 

229.  The  point  M  is  -2§";  o";  -2".  The  point  N  is  \\"\  -4";  -2". 
The  line  MN  is  the  axis  of  a  prism  whose  base  in  V  is  a  rectangle  2"  by  1" . 
The  point  O  is  3 a";  -2";  q" .  The  point  Q  is  o";  -^\"\  -3^-  The  line 
OQ  is  the  axis  of  an  hexagonal  pyramid  whose  base  in  H  has  \"  sides.  Draw 
a  plan  and  elevation  showing  the  intersection. 

230.  The  point  A  (|";  —4";  —25")  is  the  apex  of  a  pyramid  whose  base 
is  CDEF.  The  point  C  is  -2H";  -i^";o.  The  point  D  is  -\h"y 
-2")  o".  The  point  E  is  -A";  -|";  o".  The  point  F  is  -2";  o";  o". 
The  point  B  (—2";  —2^";  —li^")  is  the  apex  of  a  second  pyramid  whose 
base  is  MNOP.  The  point  M  is  H";  o'';  o".  The  point  N  is  2A";  -ih"; 
o".  The  point  O  is  W\  o";  o".  The  point  P  is  2^j/')  o"\  o".  Draw  a 
plan  and  elevation  of  these  two  pyramids  showing  their  intersection.  Make 
a  development  of  the  pyramid  A-CDEF  with  this  line  of  intersection  traced 
on  it. 

231.  The  axis  of  a  6"  square  timber  pierces  V  at  o";  o";  — 18",  is  parallel 
to  H,  and  inclines  60  degrees  to  V.  The  axis  of  a  second  timber  of  the  same 
size  pierces  H  at  +6";  —18';  o",  is  parallel  to  V,  and  inclines  60  degrees  to 
H.  Draw  a  plan  and  elevation  of  these  two  timbers  and  show  also  a  plan  and 
elevation  of  the  shortest  6"  square  timber  which  can  be  fitted  between  them. 


142  ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 

PROBLEMS   ON  THE  INTERSECTION   OF   CYLINDERS 

232.  The  axis  of  a  cylinder  whose  base  in  H  is  a  2"  circle  pierces  H  at 
o;  _ir".o'\  A  second  point  on  this  axis  is  at  2";  — sH";  —2".  The 
axis  of  a  second  cylinder  whose  base  in  H  is  a  li"  circle  pierces  H  at  2\"\ 

—  1 1";  o".  A  second  point  on  the  axis  of  this  cylinder  is  at  i"\  —\)!'\  —2\", 
Draw  a  plan  and  elevation  of  these  two  intersecting  cylinders  and  show  the 
plan  and  elevation  of  their  curve  of  intersection. 

233.  The  axis  of  a  cylinder  whose  base  in  H  is  a  2"  circle  pierces  H  at 
o";  —  i^'';  o".  A  second  point  on  this  axis  is  at  2";  —  2H";  —2".  The 
axis  of  a  second  cyUnder  whose  base  in  H  is  a  i\"  circle  pierces  H  at  3I"; 

—  \"\  o" .  A  second  point  on  this  axis  is  at  2";  —2\"\  —2\".  Show  the 
plan  and  elevation  of  the  curve  of  intersection  of  these  two  cylinders. 

234.  The  axis  of  a  cylinder  whose  base  in  V  is  a  2\"  circle  pierces  V  at 
o";  o";  — 3I".  The  axis  of  a  second  cylinder  whose  base  is  a  \\"  circle 
pierces  H  at  \\"\  —4'';  o" .    These  two  axes  intersect  at  the  point  2^'\ 

—  2\";  —2^".    Draw  a  plan  and  elevation  of  the  curve  of  intersection. 

235.  The  axis  of  a  cylinder  whose  base  in  V  is  a  i:^''  circle  pierces  V  at 
o";  o";  1 1";  a  second  point  on  this  axis  is  at  2";  —2,2")  —2\".  The  axis 
of  a  second  cylinder  whose  base  in  H  is  a  1 1"  circle  pierces  H  at  the  point 
2\"\  —  i;  o".  A  second  point  on  this  axis  is  at  o";  —3!'';  —41".  Draw 
the  plan  and  elevation  of  the  two  cylinders  showing  their  curve  of  inter- 
section. 

236.  The  axis  of  a  12"  pipe  lies  parallel  to  H  and  36''  below  H;  it  in- 
clines 60  degrees  to  V.  The  axis  of  a  second  12"  pipe  is  parallel  to  the  G.  L. 
2^"  back  of  V  and  60"  below  H.  Draw  the  plan  and  elevation  of  these  two 
pipes  and  make  the  pattern  for  the  shortest  8"  pipe  which  will  connect  them. 

PROBLEMS   ON  THE  INTERSECTION  OF   CONES 

237.  The  center  of  the  base  of  a  cone  is  at  o";  —3'';  o".  The  base  is  a 
circle  3"  in  diameter.  The  apex  of  the  cone  is  at  ^\";  —51";  —sV-  The 
center  of  the  base  of  a  second  cone  is  at  3^'';  —if'';  o";  the  base  is  a  circle 
2"  in  diameter.  The  apex  of  this  cone  is  at  2";  —51'';  —4''.  Draw  a  plan 
and  elevation  of  the  two  cones  showing  their  curve  of  intersection. 

238.  The  center  of  the  base  of  a  cone  is  at  o";  —2>W',  —if"-  The  base 
is  a  3"  circle  whose  plane  is  parallel  to  V.  The  apex  of  this  cone  is  at  45"; 
-■h")  -5f'.  The  base  of  a  second  cone  is  at  3";  -3H";  -2".  The 
base  of  this  cone  is  a  2  \"  circle  whose  plane  is  parallel  to  V.  The  apex  is  at 
the  point  o";  —^"]  —^l".  Draw  a  plan  and  elevation  of  these  two  cones 
and  show  their  curve  of  intersection. 

239.  The  base  of  a  cone  in  V  has  its  center  at  the  point  o";  o"\  — 1|";  the 
diameter  of  the  base  is  2^".  The  apex  of  this  cone  is  at  2|";  —3!";  —  2!". 
The  base  of  a  second  cone  in  H  has  its  center  at  3";  —i\"\  o"\  the  base  is 


INTERSECTION  OF  SURFACES  143 

The  apex  of  this  cone  is  at  f";  — 1|";  —  3I''.    Draw  a 
plan  and  elevation  of  the  cones  showing  their  curve  of  intersection. 

240.  The  base  of  a  cone  in  H  has  its  center  at  o"\  — 1|";  o";  the  diam- 
eter of  the  base  is  2 1".  The  apex  of  this  cone  is  at  4I'' ;  —  3  f " ;  —  4i".  The 
base  of  a  second  cone  in  V  is  at  ^\";  o";  —  if";  the  base  is  2\"  in  diameter. 
The  apex  of  this  cone  is  at  —  |";  —5";  —2^".  Show  a  plan  and  elevation 
of  the  two  cones  and  find  their  curve  of  intersection, 

241.  The  center  of  the  base  of  a  right  cone  is  2I"  behind  V,  and  its  apex 
is  3 1"  below  H.  The  cone  is  perpendicular  to  H  and  its  base  in  H  is  3I". 
A  second  right  cone,  which  is  perpendicular  to  V,  has  the  center  of  its  base 
i\"  below  H  in  V,  and  its  apex  is  4^"  behind  V.  The  base  of  the  second 
cone  is  a  circle  2\"  in  diameter.  Draw  a  plan,  elevation,  and  end  view  of 
the  two  cones,  and  show  three  views  of  their  curve  of  intersection. 

242.  The  center  of  a  2^"  circle  whose  plane  is  parallel  to  V  is  at  o"; 
—3";  —2".    This  circle  is  the  base  of  a  right  cone  whose  apex  is  at  o"; 

—  h"',  —2".     The  center  of  the  base  of  a  second  cone  is  at  2//';  ~^^'\ 

—  2\"\  the  apex  of  this  cone  is  at  —2\"\  —2I";  —  f".  The  base  of  this 
second  cone  is  a  2 1"  circle  lying  in  the  plane  1^";  300;  270.  Show  the  plan 
and  elevation  of  these  cones  and  their  curve  of  intersection. 


PROBLEMS   ON  THE  INTERSECTION  OF  CYLINDERS  AND 

CONES 

243.  The  center  of  the  base  of  a  cylinder  lies  in  H  at  the  point  o"; 

—  2";  o";  the  base  is  a  circle  3"  in  diameter.  A  second  point  on  the  axis 
of  the  cyHnder  is  at  3";  —35";  —  if.  The  center  of  the  base  of  a  cone  lies 
in  H  at  the  point  4^";  —  li";  o";  the  diameter  of  this  base  is  2".  The 
apex  of  this  cone  is  at  I";  —4 1";  —  3i".  Draw  a  plan  and  elevation  of  the 
cone  and  cylinder  showing  the  curve  of  intersection. 

244.  The  base  of  a  cylinder  has  its  center  in  H  at  o";  —iV;  o";  the 
diameter  of  the  base  is  2".  A  second  point  on  the  axis  of  this  cylinder  is  at 
2";  — 3I";  —2".  The  center  of  the  base  of  a  cone  lies  in  H  at  the  point 
3";  —  1 5";  o";  the  diameter  of  this  base  is  if".  The  apex  of  the  cone  is  at 
4";  ~3l";  —3 1"-  Draw  the  plan  and  elevation  of  the  cylinder  and  cone 
and  show  their  curve  of  intersection. 

245.  The  base  of  a  cylinder  has  its  center  in  V  at  o";  o";  —  ig";  the 
diameter  of  this  base  is  i  ^".  A  second  point  on  the  axis  of  the  cylinder  is  at 
2";  -3tV';  -2i".  The  center  of  the  base  of  a  cone  lies  in  Hat  3!";  -itV; 
o";  the  diameter  of  the  base  is  2^".    The  apex  of  the  cone  is  at  the  point 

—  I" ;  —  2" ;  —  3 1".  Draw  a  plan  and  elevation  of  the  two  surfaces  and  show 
the  curve  of  intersection. 

246.  A  cylinder  3"  in  diameter  lies  parallel  to  H  and  V  with  its  axis  2" 
back  of  V  and  3"  below  H.  The  apex  of  a  right  cone  coincides  with  this 
axis  of  the  cylinder  and  the  axis  of  the  cone  inclines  60  degrees  to  both  H 


144  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

and  V.  A  right  section  of  the  cone  2"  from  the  apex  is  2"  in  diameter. 
Draw  a  plan,  elevation,  and  end  view  of  the  cone  and  cylinder  and  show 
their  curve  of  intersection. 

247.  The  axis  of  a  right  cone  pierces  V  \\"  above  H  and  inclines  45  de- 
grees to  V;  the  apex  is  4''  above  H.  Between  these  two  points  the  axis  of 
the  cone  is  4''  long,  and  a  right  section  of  the  cone  3''  from  the  apex  is  a  circle 
\\"  in  diameter.  At  a  point  2"  from  the  apex  the  axis  of  the  cone  is  cut 
by  the  axis  of  a  right  circular  cylinder  2"  in  diameter  whose  axis  is  parallel 
to  H  and  V.  Draw  three  views  of  these  two  surfaces  showing  the  curve  of 
intersection. 


CHAPTER  XVI 


SURFACES   OF   REVOLUTION 

114.  A  Surface  of  Revolution  is  represented  in  a  drawing  by 
the  projections  of  its  meridian  curves,  or  sections.  These  me- 
ridian curves  are  always  taken 
in  planes  through  the  axis 
parallel  to  the  planes  of  pro- 
jection. Thus,  in  Fig.  114, 
the  plan  view  of  the  sphere  is 
the  projection  of  the  curve 
cut  from  the  sphere  by  the 
meridian  plane  AB;  and  the 
elevation  of  the  sphere  is  the 
elevation  of  the  curve  cut 
from  the  sphere  by  the  me- 
ridian plane  CD. 

115.  To  assume  a  point  on 
the  surface  of  a  surface  of 
revolution  one  projection  may 
be  assumed,  as  x  in  Fig. 
114,  and  the  other  projection 
found  by  passing  a  meridian 
plane  MN  through  the  as- 
sumed point  and  revolving 
this  plane  into  coincidence 
with  the  meridian  plane  CD. 
In  this  position  the  point  X 
will  be  projected  horizontally 
at  X/,  and  vertically  at  x"  on  the  meridian  curve.  The  plane 
MN  may  now  be  revolved  back  to  its  original  position  and  the 
projection  of  point  X  on  the  surface  will  be  found  at  x  and  x'. 

145 


Fig.  114. 


146  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

It  will  be  noted  that  the  point  projected  on  H  at  x  may  have 
two  elevations  and  which  of  these  is  to  be  taken  depends  upon 
which  side  of  the  sphere  the  point  is  assumed  originally. 

116.  In  finding  the  intersections  cut  from  surfaces  of  revolu- 
tion by  other  surfaces  the  problem  will  be  greatly  simpHfied  if 
the  following  rule  is  observed: 

Pass  the  auxiliary  planes  through  the  given  surface  in  such  a 
direction  that  only  the  simplest  and  most  easily  located  curves  and 
elements  of  the  surfaces  are  cut  out. 

Since  surfaces  of  revolution  have  circular  sections  in  planes 
perpendicular  to  the  axes  of  revolution  it  will  be  obvious  that, 
where  expedient,  the  cutting  planes  should  be  passed  in  this 
direction. 

117.  Proposition  35.  Given  any  surface  of  revolution  to  find 
its  intersection  with  any  plane. 

Discussion.  Pass  auxiliary  planes  through  the  surface  per- 
pendicular to  the  axes  of  revolution.  These  planes  cut  circles 
from  the  surface  and  lines  from  the  given  plane.  Since  each 
circle  and  line  lies  in  a  common  plane  they  will  intersect  in  points 
common  to  the  given  surfaces,  hence  on  the  required  intersection. 

Construction.  In  Fig.  115  the  given  surface  is  an  ellipsoid 
with  its  axis  of  revolution  perpendicular  to  H.  It  is  cut  by  the 
plane  sSs'.  To  find  the  required  intersection  pass  planes  through 
the  plane  and  ellipsoid  perpendicular  to  the  axis  or  parallel  to  H, 
as  R.  The  plane  R  cuts  the  line  MN  from  the  plane  S  and  the 
circle,  whose  diameter  is  AB,  from  the  ellipsoid.  This  line  inter- 
sects the  circle  at  X  and  Y,  two  points  on  the  required  curve. 
In  like  manner  plane  V  cuts  the  circle  whose  diameter  is  CD  from 
the  surface  and  from  the  plane  cuts  the  line  OP.  These  intersect 
at  0  and  P,  two  more  points  on  the  required  curve;  continuing 
this  process  enough  points  may  be  found  to  locate  the  curve.  If 
the  true  size  of  the  curve  be  required  it  may  be  found  by  re- 
volving it  into  H  or  V  about  the  corresponding  trace  of  plane  S. 

118.  Proposition  36.  Given  any  surface  of  revolution  to  find 
its  intersection  with  any  cylinder. 

Discussion.  Pass  auxiliary  planes  through  the  surfaces  so 
that  circles  are  cut  from  the  surface  of  revolution  and  elements 


SURFACES  OF  REVOLUTION 


147 


from  the  cylinder.  These  elements  cut  the  circles  in  points 
common  to  both  surfaces  and  are  points  of  the  required  inter- 
section. 


Construction.  In  Fig.  116  the  cylinder  has  a  circular  base  in 
H  and  intersects  a  semicircular  pipe  bend  (annular  torus) .  The 
auxiHary  plane  R  cuts  from  the  surface  of  the  torus  a  circle  which 


148 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


is  the  elevation  of  the  surface,  and  from  the  cylinder  the  limiting 
elements  of  the  elevation.  The  elements  cut  this  circle  at  A  and 
F,  or  two  points  on  the  required  curve  of  intersection.  In  like 
manner  auxiliary  plane  S  cuts  a  smaller  circle  from  the  torus  and 


Fig.  ii6. 


two  elements  from  the  cylinder.  These  intersect  at  E  and  B, 
two  more  points  on  the  curve.  Auxiliary  plane  T  locates  points 
D  and  C,  and  in  this  manner  the  entire  curve  may  be  located. 

119.  Proposition  37.     Given  any  surface  of  revolution  to  J&nd 
its  intersection  with  a  cone. 


SURFACES  OF  REVOLUTION 


149 


Discussion.     See  Article  116. 

Construction.  In  Fig.  117  is  a  cylinder  of  revolution  with  a 
hemispherical  top,  intersected  by  a  cone  with  a  circular  base. 
From  a  study  of  the  figure  it  will  appear  that  the  auxiUary 
planes  should  be  passed  parallel  to  H,  for  in  this  direction  each 


Fig.  117. 


plane  cuts  a  circle  from  each  surface.  Thus,  plane  R  cuts  from 
the  cone  a  circle  whose  radius  is  y'l  and  from  the  other  surface  a 
circle  whose  radius  is  x'2.  These  two  circles  intersect  at  A  and 
B  two  points  on  the  required  curve.  By  passing  other  planes 
parallel  to  R  the  curve  of  intersection  may  be  completely  deter- 
mined. 


I50 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


120.  Proposition  38.  Given  any  two  surfaces  of  revolution 
whose  axes  intersect  to  find  the  curve  of  intersection  of  the 
surface. 


1             ' ' 

^■JX 

A        \  1    \ 

Fig. 


Discussion.  If,  with  the  intersection  of  the  axes  of  the  given 
surfaces  as  a  center,  auxiliary  spheres  be  passed  through  the 
surfaces  they  will  cut  circles  from  each  surface.  These  circles 
intersect  in  points  on  the  curve  of  intersection. 


SURFACES  OF  REVOLUTION 


151 


Construction.  In  Fig.  118  the  sphere  is  cut  by  a  cone.  The 
axes  intersect  at  O.  With  this  point  as  a  center  describe  a  sphere 
as  ABC.  This  sphere  cuts  from  the  given  sphere  a  circle  verti- 
cally projected  as  a  line  at  a'b'  and  from  the  cone  another  circle 
vertically  projected  as  a  line  at  m'n'.  These 
two  circles  intersect  at  two  points,  both  of 
which  are  vertically  projected  at  x'  and 
horizontally  projected  on  the  plan  view  of 
circle  AB  at  x  and  y.  By  passing  other 
spheres  and  proceeding  in  the  same  manner 
the  curve  of  intersection  may  be  completely 
determined. 

121.   Proposition  39.     To  develop  any  sur- 
face of  revolution. 

Discussion.      Theoretically,    surfaces    of 
revolution  are  not  capable  of  development 
but,  practically,  such  surfaces  are  developed 
by    methods  .  which    give    approximations, 
which  are  near  enough  for  commercial  pur- 
poses.    Examples  of  this  are  shown  in  the 
following  developments  of  the  sphere.     In 
the  first  method  the  surface  of  the  sphere 
is  divided  by  meridian  planes  into  gores  as 
shown  in  Fig.  119.     In  the  second  method, 
Fig.  120,  the  surface  of  the 
sphere  is  divided  into  zones 
by    making    each    zone    a 
frustum  of  some  cone  which 
nearly    coincides   with    the 
surface  of  the  sphere.     Of- 
ten these  two  methods  are 
combined  in  laying  out  the  bottoms  of  hemispherical  tanks. 

Construction.  Gore  Method.  Divide  the  surface  into  gores 
by  meridian  planes,  keeping  in  mind  the  fact  that  the  narrower 
the  gores  are  made  the  more  nearly  will  the  development  ap- 
proach the  true  surface  of  a  sphere.  In  Fig.  119  the  hemisphere 
is  divided  into  16  equal  gores.     To  lay  out  each  gore  divide  the 


152 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


sphere  as  shown  in  the  figure  by  horizontal  planes  1-2-3-4,  etc., 


and  lay  off  the  distances  1-2;    2-3; 


3-4,  etc., 


in  the  line  1-7, 


thus  making  the  line  1-7  equal  to  J  the  great  circle  of  the  sphere. 
Through  the  point  i,  2,  3,  4,  etc.,  thus  located  draw  horizontal 
lines;   these  will  be  the  developed  circles  cut  by  the  horizontal 


SURFACES  OF  REVOLUTION  153 

planes  through  i,  2,  3,  4,  etc.  On  each  of  these  horizontal  lines 
lay  off  the  distances  between  the  meridian  planes.  Thus,  at  the 
point  I  lay  off  ab  equal  to  the  arc  ab;  at  point  2  lay  off  the 
length  of  arc  between  the  meridian  planes  in  the  plane  2. 
Sixteen  gores,  arranged  as  shown  in  the  figure,  developed  by 
this  method  will  give  an  approximate  pattern  of  the  hemisphere. 

Construction.  Zone  Method.  Divide  the  surface  of  the 
hemisphere  in  Fig.  120  into  zones  by  making  each  zone  the  frus- 
tum of  a  cone  whose  apex  is  on  the  axis  of  the  hemisphere  and 
whose  elements  are  tangent  to  the  hemisphere.  Develop  these 
frusta  by  any  convenient  method  of  laying  out  a  cone  and 
arrange  them  as  shown ;  the  result  will  approximate  very  closely 
a  development  of  a  sphere. 

As  in  the  gore  method,  it  should  be  remembered  that  the  more 
zones  used  the  nearer  the  result  will  approach  the  surface  of  a 
hemisphere. 

PROBLEMS   ON  SURFACES   OF  REVOLUTION 

248.  An  ellipsoid  of  revolution  is  generated  by  revolving  an  ellipse  2"  by 
3''  about  its  major  axis.  Cut  this  surface  by  a  plane  oblique  to  both  H  and 
V  and  find  the  plan  and  elevation  of  the  curve  of  intersection  and  its  true 
size. 

249.  An  annular  torus  is  generated  by  revolving  a  i"  circle  about  an 
axis  lying  in  the  plane  of  this  circle  and  i  \"  from  its  center.  Cut  this  torus 
by  a  plane  oblique  to  both  H  and  V  and  find  the  plan,  elevation,  and  true 
size  of  the  curve  of  intersection. 

250.  The  center  of  a  hemisphere  is  at  o";  —3";  o".  The  center  of  the 
base  of  a  cylinder  is  at  5";  —3";  o".  The  radius  of  the  hemisphere  is  i|", 
and  the  base  of  the  cylinder  is  an  ellipse  lying  in  H  with  its  2"  major  axis 
perpendicular  to  the  G.  L.  and  its  \\"  minor  axis  parallel  to  the  G.  L.  The 
axis  of  the  cylinder  is  parallel  to  V  and  inclines  15  degrees  to  H.  Draw  a 
plan  and  elevation  of  these  two  surfaces  showing  the  curve  of  intersection. 

251.  A  tank  with  a  hemispherical  bottom  is  drained  by  a  pipe  which 
runs  off  at  an  angle  of  30  degrees  to  the  horizontal.  The  tank  is  72''  in 
diameter,  the  pipe  is  d"  in  diameter,  and  the  axis  of  the  pipe  pierces  the 
bottom  of  the  tank  at  a  point  on  its  surface  directly  below  the  center 
of  its  hemispherical  bottom.  Draw  a  plan  and  elevation  of  the  tank  and 
pipe  showing  the  curve  of  the  joint  between  them. 

252.  The  axis  of  an  ellipsoid  of  revolution  is  perpendicular  to  H  3" 
behind  V.    The  surface  is  generated  by  the  revolution  of  an  ellipse  3''  by  2", 


154  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

the  2"  dimension  being  parallel  to  H  and  the  upper  end  of  the  3"  axis  being 
•l\"  below  H.  The  axis  of  a  right  cone  is  parallel  to  the  G.  L.  3"  below  H 
and  2!'  behind  V.  The  apex  of  the  cone  is  2''  to  the  right  of  the  axis  of  the 
ellipsoid  and  the  center  of  the  base  Hes  in  a  profile  plane  2"  to  the  left  of  the 
axis.  The  base  of  the  cone  is  a  circle  2\"  in  diameter  lying  in  this  profile 
plane.  Draw  a  plan  and  elevation  of  these  two  surfaces  and  show  their 
curve  of  intersection. 

253.  In  the  Pierce-Arrow  automobile  the  headlights  —  which  are  pa- 
raboloids of  revolution  —  are  set  into  the  mud  guards  of  the  front  wheels. 
Assuming  the  necessary  dimensions  find  the  shape  of  the  hole  which  will 
have  to  be  cut  into  the  cylindrical  mud  guard  to  receive  the  lamp. 

254.  Make  the  patterns  for  the  bottom  of  a  tank  which  is  a  hemisphere 
of  42"  radius. 

255.  Show  how  to  cut  the  tin  to  cover  a  dome  which  is  a  semi-ellipsoid 
in  form  with  axes  84"  by  to" . 

256.  Draw  three  views  of  a  U.  S.  Standard  hexagonal  nut  for  a  2"  bolt. 
The  nut  has  a  spherical  chamfer. 


CHAPTER  XVII 
WARPED   SURFACES 

122.  A  Warped  Surface  is  represented  by  the  projections  of 
its  directrices  and  a  sufficient  number  of  positions  of  its  genera- 
trix to  show  the  character  of  the  surface,  and  to  indicate  how  the 
generating  line  moves.  Thus,  in  Fig.  71,  the  projections  of  the 
directrices  AB  and  CD  are  shown  and  a  sufficient  number  of 
positions  of  the  generatrix  AX  are  given  to  indicate  the  character 
of  the  surface  and  that  AX  remains  parallel  to  the  H  plane,  which 
in  this  case  is  the  plane  director. 

123.  In  addition  to  the  warped  surfaces  which  are  unclassified 
there  are  three  general  kinds  of  warped  surfaces : 

1.  Warped  surfaces  with  two  linear  directrices  and  a  plane 
director. 

2.  Helicoidal  surfaces. 

3.  Warped  surfaces  with  three  linear  directrices. 

Some  of  these  surfaces  have  commercial  applications  and  such 
as  these  will  be  studied  in  detail  but  the  majority  of  warped 
surfaces,  owing  to  their  complex  nature  and  undevelopable 
character,  are  not  used  in  practical  work  and  possess  only  mathe- 
matical interest.  The  following  table  gives  a  classification  of 
warped  surfaces  with  the  names  of  some  of  the  more  important 
ones. 

CLASSIFICATION  OF  WARPED   SURFACES 

I.  Linear  directrices,    both        Hyperbolic  paraboloid 
I.    Warped  surfaces  straight  lines 

with  two  linear     2.  Linear     directrices,    one        Conoid 
directrices  and  straight  and  one  curved 

a  plane  director    ^    linear   directrices,   both        Cylindroid 
curved  lines 

15s 


156 


ESSENTIALS   OF   DESCRIPTIVE   GEOMETRY 


2.  Warped  surfaces 
with  a  helical 
directrix 


3.  Warped  surfaces 
with  three  linear 
directrices 


1.  Right  helicoid  Generatrix  perpendicular  to  the  axis 

2.  Oblique  helicoid  Generatrix  oblique  to  axis 

3.  Convolute    (de-  Not  a  warped  surface  but  a  single 

velopable,  heli-  curved  surface  to  which  the  heli- 
coid) coid  approaches  as  a  limit 

I.  Linear    directrices,  Hyperboloid  of  one   nappe.    Also 


3- 


all  straight  lines 

Two  linear  direc- 
trices, one  straight 
and  one  curved 

One  linear  directrix 
straight  and  two 
curved 


4.  Linear    directrices, 
all  curved  lines 


a  surface  of  revolution 


Warped  arch 
Warped  cone 


4.  Warped  surfaces 
not  classified 


Serpentine 
Conical  spring,  etc. 


124.  To  assume  a  point  on  a  warped  surface  one  of  two 

methods  may  be  used.  In  case  it  is  possible  to  draw  the  pro- 
jection of  an  element  through  the  assumed  projection  of  the 
point,  as  in  Fig.  121,  this  may  be  done  and  the  other  projection 
of  the  point  may  be  found  on  the  corresponding  projection  of 
the  element.  In  Fig.  121,  x'  was  assumed  and  through  it  was 
drawn  the  projection  p'q'  of  an  element  of  the  surface  passing 
through  that  point.  The  H  projection  of  this  element  is,  of 
course,  pq,  and  on  this  projection  the  other  projection  x  may 
be  located. 

In  Fig.  122  this  method  is  not  applicable  since  through  the 
assumed  projection  of  the  point  x  it  is  not  convenient  to  draw  the 
projection  of  the  element  containing  it.  To  find  x'  in  this  case, 
pass  a  plane  R  through  the  point  perpendicular  to  H.  This  plane 
cuts  from  the  warped  surface  the  line  OPQ,  and  obviously  x'  will 
be  found  located  on  its  projection  o'p'q'. 

125.  Of  warped  surfaces  having  a  plane  director  Fig.  71  is  an 
example.  This  surface  has  two  straight  line  directrices  and  is 
a  common  example  of  warping.  This  surface  is  called  an  hyper- 
bolic paraboloid  because  one  set  of  planes  will  cut  hyperbolas 


WARPED   SURFACES 


157 


Fig.  122. 


158 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


from  the  surface  while  another  set  perpendicular  to  the  first  set 
will  cut  parabolas.     (See  Fig.  123.)     As  practical  examples  of 

this  surface  the  pilot  on  a 
locomotive  and  the  bow  of 
a  ship  may  be  cited. 

If  one  directrix  is  a  curved 
line    and    one    remains    a 
straight  line  the  surface  be- 
comes   a    conoid.    An    ex- 
ample   of    such    a    surface 
may    be    found    in    chisels 
Fig.  123.  —  An  hyperbolic  paraboloid.    Note  shaped  from  round  bar  stock 
that  planes  parallel   to   the  base  of  the    and    in    the    bows   of    ships, 
model  wUl  cut  out  hyperbolas,  and  that   pjg     jgl    shows    an   obUque 

conoid,  and  Fig.  1 24  shows  a 

stringmodelof  a  right  conoid. 

become    curved    lines    the    surface 


planes  parallel  to  its  ends  will  cut   out 
parabolas. 


Where  both  directrices 
becomes  a  cylindroid. 
Fig.  122  shows  such  a 
surface  and  illustrates 
the  skewed  arch,  one  of 
the  useful  applications 
of  the  surface.  The 
<:ylindroid  is  also  made 
use  of  in  buildings  for 
connecting  arched  pas- 
sageways at  different  ele- 
vations. 

Many  interesting  ap- 
plications of  these  and 
other  warped  surfaces 
may  be  seen  in  the 
bodies,  mud  guards,  and 
other  portions  of  modern 
automobiles. 

126.   Proposition  40.     Given   the  directrices  and   the  plane 
director  to  draw  the  plan  and  elevation  of  a  warped  surface. 


Fig.  124.  —  A  string  model  of  a  right  conoid. 
The  base  of  this  surface  is  a  circle  which  is  also 
the  curved  directrix;  the  bar  at  the  top  is  the 
straight  line  directrix.  The  plane  director  is 
parallel  to  the  ends  of  the  box. 


WARPED   SURFACES  1 59 

Discussion.  The  directrices  may  both  be  straight  lines,  or 
may  both  be  curved  lines,  or  one  may  be  straight  and  one  curved. 
The  character  of  the  surface  will,  of  course,  be  determined  by 
the  character  of  these  directrices  as  described  in  Article  125,  but 
the  method  of  drawing  will  be  the  same  for  all  cases. 

Construction.  In  Fig.  122  let  AB  and  CD  be  the  two  given 
directrices,  in  this  case  both  being  curviUnear.  Let  the  H  plane 
be  the  plane  director.  From  any  point.  A,  on  one  directrix 
draw  a  line,  as  AC,  parallel  to  the  plane  director,  H.  This 
line  intersects  the  other  given  directrix  at  C.  Therefore,  since 
AC  touches  both  directrices  and  is  parallel  to  the  plane  director 
it  will  be  an  element  of  the  surface.  In  like  manner  other 
elements  may  be  drawn  and  the  plan  and  elevation  outlined. 
In  commercial  drafting  only  the  limiting  elements,  or  those  ele- 
ments which  define  the  contour  of  the  surface,  are  shown. 

In  Fig.  121,  the  obhque  conoid,  the  method  of  construction 
is  the  same.  In  this  case  the  curviUnear  directrix  is  the  circle 
MN  and  the  rectilinear  directrix  is  AB,  and  the  plane  director 
is  an  oblique  plane  rRr'.  The  elements  must,  of  course,  touch 
both  MN  and  AB  and  be  parallel  to  plane  R. 

127.  A  helicoidal  surface  is  generated  by  a  line  moving  uni- 
formly about  and  along  an  axis,  making  a  constant  angle  with 
it.  If  the  generatrix  is  of  definite  length  it  is  obvious  that 
during  the  generation  of  such  a  surface  one  end  of  the  genera- 
trix —  that  touching  the  axis  —  will  describe  a  straight  line, 
while  the  other  end  will  follow  a  path  in  space  which  is  a  curve 
of  double  curvature  called  a  helix.  It  must  be  apparent  that  all 
points  between  these  extreme  points  of  the  generatrix  will  also 
describe  hehces  of  sizes  which  will  vary  according  to  the  position 
of  the  point  on  the  generatrix.  The  surface  of  the  hehcoid,  then, 
may  be  said  to  be  made  up  of  a  series  of  concentric  hehces,  and 
in  order  to  study  such  a  surface  properly  the  nature  of  the  helix 
must  first  be  considered. 

128.  The  helix  is  a  space  curve  generated  by  moving  a  point 
around  and  along  an  axis  at  a  uniform  rate.  The  distance  from 
the  axis  may  be  constant  giving  a  cylindrical  helix  (Fig.  125),  or 
it  may  vary  uniformly  giving  a  conical  helix  (Fig.  126).     In 


i6o 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


either  of  these  cases  the  distance  the  generating  point  moves  in 
the  direction  of  the  axis  during  one  revolution  is  called  the  pitch. 
Practical  examples  of  such  curves  may  be  found  in  the  center 
lines  of  cylindrical  and  conical  springs  (Figs.  127  and  128). 


Fig.  125. 


Fig.  126. 


129.  To  construct  a  cylindrical  helix  it  is  necessary  to  know 
the  pitch  and  diameter  of  the  cylinder  on  which  the  helix  may  be 
wrapped.  In  Fig.  129  the  pitch  and  diameter  of  the  cylinder  are 
given.    The  plan  view  of  the  curve  will  be,  of  course,  a  circle  and 


WARPED   SURFACES 


i6i 


may  be  drawn  at  once.  To  find  the  elevation,  divide  the  pitch 
distance  into  any  number  of  convenient  parts,  say  24,  and  divide 
the  circle  of  the  H  projection  into  the  same  nimiber  of  parts. 
Let  the  generating  point  start  in  H  at  point  o.  In  moving  from 
o  to  I  the  point  makes  one  twenty-fourth  of  a  complete  turn 


Fig.  127. 


about  the  axis,  so  it  therefore  advances  in  the  direction  of  the 
axis  one  twenty-fourth  of  the  pitch;  i'  then  is  the  elevation  of 
its  second  position.  In  Hke  manner  all  twenty-four  positions 
may  be  located  and  the  curve  plotted. 

In  drawing  a  conical  helix  the  same  principle  appHes  except 


l62 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


that  it  must  be  remembered  that  the  H  projection  will 
spiral  and  not  a  circle  and  that,  since  the  helix  is  on  the 
of  a  cone,  the  distance  of  the  generating  point  from  the  axis 
varies  directly  with  the  distance  from  its  first  position.    Fig.  126 


ill  be  a  I 
surface  I 


Fig.  128. 


Fig.  129. 


shows  the  construction  of  a  conical  helix  and  the  method  of 
finding  its  projections. 

130.  To  Draw  a  Tangent  to  a  Helix.  If  in  any  given  helix  the 
generating  point  starts  from  H,  as  in  Fig.  125,  it  will  be  found 
that  the  distance  from  any  position  of  the  point  on  the  curve 
back  to  H  along  the  path  of  the  helix  is  equal  to  the  distance  from 


WARPED   SURFACES  1 63 

that  same  point  back  to  H  along  the  path  of  the  tangent  to  the 
helix  at  that  point.  This  is  true  because  the  curve  and  its  tan- 
gent make  the  same  angle  with  H,  and  it  may  be  proved  by 
wrapping  a  string  around  a  cyHnder  in  the  form  of  a  helix.  Now 
if  a  portion  of  the  string  be  unwound  from  the  cylinder  it  will  be 
seen  that  the  unwrapped  portion  is  tangent  to  the  portion  of  the 
helix  still  wrapped  on  the  cylinder,  and  that  the  end  of  the  string 
always  remains  in  the  same  plane.  This  fact  is  made  use  of  in 
drawing  a  tangent  to  a  helix.  In  Fig.  125  draw  px  perpendicular 
to  op,  and  make  px  equal  to  p-1.  This  Une  will  be  the  H  pro- 
jection of  the  tangent  to  the  helix  and  x  will  be  the  point  where 
the  tangent  pierces  H,  because  px  is  the  plan  of  a  tangent  line 
equal  in  length  to  the  length  of  the  helix  from  H  to  P.  p'x'  then 
will  be  the  elevation  of  the  tangent  and  will  be  tangent  to  the 
elevation  of  the  helix. 

131.  Proposition  41.  Given  the  axis,  the  generatrix,  and  the 
pitch  to  draw  the  plan  and  elevation  of  a  helicoidal  surface. 

Discussion.  A  helicoidal  surface  may  be  best  represented  by 
showing  a  plan  and  elevation  of  its  axis,  its  base  in  H  or  V,  and 
a  sufficient  number  of  elements  to  define  the  contour  of  the  sur- 
face. For  use  in  solving  problems  in  connection  with  this  surface 
it  is  convenient  to  have  also  the  plan  and  elevation  of  the  helix 
described  by  the  end  of  the  generating  line. 

Construction.  In  Fig.  130  let  AB  be  the  generating  line  in  its 
first  position.  Let  the  axis  be  the  line  through  A  perpendicular 
to  H,  and  let  the  distance  that  the  point  B  rises  during  one 
revolution  of  AB  about  the  axis  be  equal  to  twice  the  distance 
from  12'  to  the  G.  L.  Draw  the  helix  described  by  point  B 
according  to  the  directions  in  Article  129.  As  the  point  B  takes 
up  the  position  on  this  helix  indicated  by  the  points  i,  2,  3,  etc., 
the  point  A  will  remain  in  contact  with  the  axis  and  will  rise  equal 
distances,  as  shown  by  a'l,  a'2,  a'3,  etc.  By  joining  the  successive 
positions  of  A  and  B  the  positions  of  the  generating  line  as  it 
moves  about  the  axis  may  be  shown.  If  the  points  where  these 
several  positions  of  the  generating  line  pierce  H  be  found,  as  at 
b,  bi,  b2,  bs,  etc.,  the  curve  of  the  base  may  be  located.  From  an 
inspection  of  the  drawing  it  will  be  seen  that  the  intersection  of 


164 


ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 


Fig.  130. 


WARPED   SURFACES 


165 


the  surface  with  H  gives  a  curve  called  the  Spiral  of  Archimedes. 
In  the  drawing  only  a  little  over  one-half  of  one  revolution  of  the 
line  AB  about  the  axis  has  been  shown. 

132.  There  are  two  general  kinds  of  helicoidal  surfaces  both 
of  which  have  many  practical  uses.  When  the  generatrix  is  in- 
chned  to  the  axis,  as  in  Fig. 
130,  the  resulting  helicoid  is 
called  an  oblique  helicoid;  and 
when  the  generatrix  is  per- 
pendicular to  the  axis,  as  in 
Fig.  134,  the  surface  is  called 
a  right  helicoid.  In  its  prac- 
tical applications  the  heli- 
coidal surface  is  not  indefinite 
in  extent  but  is  limited  usually 
by  cylinders.  Fig.  131  shows 
a  plan  and  elevation  of  an 
oblique  hcHcoidal  surface  in 
the  form  of  a  screw  thread. 
Fig.  132  shows  a  right  heU- 
coidal  surface  such  as  is  used 
in  cams.  Practical  examples 
of  hehcoids  may  be  found  in 
so-called  ''spiral"  stairways, 
screw  conveyors,  propellers, 
screw  threads,  twist  drills,  etc. 

133.  Proposition  42.  Given 
the  axis,  the  pitch,  and  gen- 
eratrix to  draw  the  plan  and 
elevation  of  a  right  helicoid.  ^^^-  ^3i- 

Discussion.  Since  nearly  all  right  helicoids  are  necessarily 
limited  in  extent  for  practical  purposes  the  method  by  which  they 
are  drawn  may  be  shown  by  a  practical  illustration.  The 
principle  involved  will  be,  of  course,  the  same  for  any  similar 
surface. 

Construction.  Let  OP  be  the  given  axis  in  Fig.  133,  and  let 
Oi  be  the  given  generatrix  which  advances  the  given  pitch  dis- 


i66 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


tance  during  one  revolution.    The  point  i  in  revolving  about  the 

axis  describes  a  circle  whose  plan  view  is  shown  at  i 13 1, 

and  moves  up  or  down  the  given  pitch  distance.     Therefore, 
divide  the  circle  into  24  equal  parts  and  the  pitch  distance  into 


Fig.  132 


the  same  number  of  parts.  As  the  point  moves  from  i  it  will 
also  advance  so  that  it  will  take  up  the  position  2,  3,  4,  etc., 
thus  describing  a  helix;  and  at  the  same  time  O  moves  along 
the  axis  so  that  the  motion  of  the  line  Oi  generates  a  right 


WARPED   SURFACES 


167 


Fig.  134. 

helicoid.  In  the  given  figure  the  square  iXYZ,  of  which  the 
line  Oi  is  one  side,  has  been  revolved,  thus  cutting  in  the  cyl- 
inder a  square  thread,  whose  faces  are  right  helicoids. 

Fig.  134  shows  a  right  helicoidal  surface  in  the  form  usually 


1 68 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


used  for  conveyors.  The  flights  are  right  helicoidal  surfaces  and 
are  mounted  on  a  shaft.  When  the  conveyor  is  enclosed  in  a 
hollow  cylinder  and  revolved  it  is  capable  of  delivering  material 
along  the  length  of  the  enclosing  casing. 

134.  When  the  axis  of  a  helicoid  becomes  a  cylinder  the 
surface  becomes  a  convolute,  and  for  this  reason  the  convolute, 
which  is  a  single  curved  surface,  is  often  called  the  developable 
helicoid.    The  helicoid  is  the  Hmiting  form  of  the  convolute, 

for  as  the  cylinder,  about 
which  the  directing  helix 
is  wrapped,  approaches  a 
straight  line  so  does  the 
convolute  surface  approach 
the  helicoid  as  a  limit.  A 
helical  convolute,  then,  is 
generated  by  a  line  moving 
tangent  to  a  given  helix. 
(See  Fig.  135.) 

135.   Proposition  43. 

Given  the  directing  helix  to 

Fig.  135.- a  string  model  of  a  developable  draw  the  plan  and  elevation 

helicoid.    The  strings  are  each  tangent  to   of  a  COnvolute  surface. 

the  helix  of  which  they  form  the  envelope,        DiscuSSion.      A  COnvolute 

and  which  may  be  seen  in  the  illustration.    ,  ,.  .     1  1       -^ 

IS  usually  represented  by  its 
base  and  the  plan  and  elevation  of  the  directing  helix,  and  a 
number  of  elements.  Since  the  surface  is  generated  by  a  line 
moving  tangent  to  the  directing  helix,  it  becomes  necessary  to 
draw  a  number  of  tangents  to  the  helix  and  find  where  they 
pierce  H  or  V  to  get  the  base. 

Construction.  In  Fig.  136,  MN  is  the  axis  of  the  given  helix 
ABCD.  By  Article  130  draw  a  number  of  tangents  to  the  helix 
and  find  where  these  tangents  pierce  H  at  i,  2,  3,  4,  etc.  The 
curve  through  these  points  will  be  the  base  in  H. 

136.  In  Fig.  137  is  shown  the  plan  and  elevation  of  a  con- 
veyor, whose  flights  are  made  developable  helicoids.  To  draw 
the  plan  and  elevation  of  such  a  surface  the  diameter  and  pitch 
of  the  smaller  helix  —  the  directrix  —  must  be  known,  and  also 


WARPED   SURFACES 


169 


the  length  of  the  generatrix  or  the  outside  diameter  of  the  con- 
veyor flights. 

Given  these  data  the  drawing  is  made  as  follows: 
Tangent  to  the  directing  helix  at  B  draw  AB.     A  is  the  point 
where  the  line,  which  is  a  position  of  the  generatrix  parallel  to 


Fig.  136. 

V,  pierces  the  limiting  outside  cylinder;  A,  then,  is  a  point  on 
the  limiting  outside  helix.  With  A  as  one  point  construct  a  helix 
with  a  pitch  equal  to  the  directing  helix  on  the  limiting  outside 
cyHnder.  If  lines  be  drawn  by  connecting  the  points  in  order, 
as  A  and  B,  etc.,  they  will  be  elements  of  the  surface.  It 
should  be  noted  that  with  these  same  given  data  another  con- 


(i7o) 


Fig.  137. 


WARPED   SURFACES  171 

volute  may  be  constructed  with  AC  as  the  generating  line,  in 
which  case  the  surface  will  slope  in  the  opposite  way  to  the  sur- 
face shown. 

137.  When  such  a  convolute,  or  developable  helicoid,  is 
developed  one  flight,  or  turn,  will  develop  into  a  flat  ring.  The 
diameter  of  the  inner  circle  of  the  ring  will  be  equal,  since  the 
tangents  of  the  inner  helix  have  a  constant  slope,  to  the  line  ed. 
This  length  is  obtained  by  drawing  a  perpendicular  b'd  to  the 
tangent  a'b',  and  from  e,  where  this  tangent  cuts  the  limiting 
element  of  the  inner  cylinder,  a  horizontal  line  ed.  The  length 
ed  is  the  graphical  solution  of  the  mathematical  equation  r  sec.^  ^, 
in  which  r  is  the  radius  of  the  inner  cyHnder  and  d  is  the  angle  the 
tangent  makes  with  the  elements  of  this  cyUnder.  The  radius 
of  the  outer  circle  of  the  ring  is  equal  to  m'o  which  length  is 
obtained  in  a  similar  manner. 

138.  The  helix  is  used  extensively  in  generating  other  warped 
surfaces  which  are  not  classified  but  which  have  considerable 
commercial  importance.  A  notable  example  of  this  is  shown  in 
the  coiled  spring.  The  surface  is  warped,  and  although  it  has 
a  helical  directrix  it  is  not  a  helicoid.  It  is  sometimes  called  a 
serpentine,  and  may  be  generated  by  moving  the  center  of  a 
sphere  of  a  given  diameter  along  a  helix.  The  method  of  draw- 
ing such  a  surface  is  shown  in  Fig.  128. 

139.  Of  warped  surfaces  of  the  third  class,  —  those  having 
three  linear  directrices,  —  comparatively  few  are  used  in  practical 
work.  These  directrices  may  all  be  curved  Hnes,  or  one  may  be 
curved  and  two  straight,  or  two  may  be  curved  and  one  straight, 
or  all  may  be  straight. 

The  peculiarity  of  this  last  surface  —  which  is  called  the  hyper- 
boloid  of  revolution  of  one  nappe  —  is  that  it  is  the  only  sur- 
face of  revolution  which  is  warped.  It  may  be  generated  by 
revolving  a  straight  line  about  another  not  in  the  same  plane 
with  it.  This  surface  is  a  surface  of  revolution,  and,  since  it  is 
possible  to  have  the  generatrix  slope  two  ways  and  generate  the 
same  surface,  the  surface  is  doubly  ruled.  (See  Fig.  138.)  It  is 
difficult  to  imagine  this  surface  as  being  also  generated  by  a  line 
moving  so  as  to  touch  three  other  straight  lines  but  that  it  can 


172 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


be  done  can  be  proved.  If  any  three  positions  of  the  generating 
line  as  it  revolves  about  the  axis  be  taken  as  directrices,  and  a 
position  of  the  generatrix  of  the  opposite  slope  be  moved  so  as 
to  touch  them,  it  will  follow  the  surface. 

The  surface  is  used  to  some  extent  in  transmitting  motion  from 
one  shaft  to  another  not  in  the  same  plane  with  it.     The  principle 

by   which    this    is   accom- 


is 


IS 

shown   in   Fig. 


plished 

139- 

If  the  line  AB  be  revolved 
about  each  shaft  it  will  gen- 
erate two  hyperboloids  hav- 
ing a  common  element,  or 
AB.  Naturally,  then,  when 
these  hyperboloids  roll  in 
their  respective  axes  they 
will  constantly  be  in  contact 
and  therefore  one  may  be 
made  to  drive  the  other. 

140.  Another  warped  sur- 
face of  this  class  having  two 
curved  directrices  and  one 

Fig.  138.  — a  model  of  an  hyperboloid  of  straight  directrix  is  some- 
one nappe  showing  by  the  position  of  the  i-ijnes  met  with  in  arches, 
strings  how  the   surface  may  be  doubly    ^r    ^ 

julg^j  If  the  two  curves  are  semi- 

circular, as  in  Fig.  140,  and 
lie  in  parallel  planes,  and  if  the  straight  line  directrix  MN 
lies  in  a  plane  perpendicular  to  the  planes  of  the  curved 
directrices  through  their  centers  the  surface  becomes  a  cow's 
horn  arch. 

A  similar  surface  of  this  same  general  character  is  the  warped 
cone,  shown  in  Fig.  141.  In  this  surface  the  curvilinear  direc- 
trices do  not  necessarily  lie  in  parallel  planes  nor  is  the  rectilinear 
directrix  in  a  plane  perpendicular  to  them.  A  conimon  appHca- 
tion  of  this  surface  is  shown  in  Fig.  141,  where  the  warped  cone 
is  used  as  a  hood.  The  method  of  construction  will  be  obvious 
from  an  inspection  of  the  figure. 


WARPED   SURFACES 


173 


Fig.  139. 


174 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


WARPED   SURFACES 


175 


141.  Proposition  44.     To  develop  a  warped  surface. 
Discussion.     While  warped  surfaces  may  not  be  truly  devel- 
oped, patterns  for  some  warped  surfaces  may  be  laid  out  which, 


a 

\ 

ill                i 
1                1 

w !   Xio'         1 

b' 

y 

P 

Fig.  141. 


when  formed,  approximate  the  original  surface  closely  enough 
for  commercial  purposes.  An  example  of  this  is  shown  in  the 
development  of  the  warped  cone  in  Fig.  142. 


176  .    ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


bo 

"t- 

0 

^w 

«=________-- 

rf5 

"co 



"bj 

-r-t 

WARPED   SURFACES  I 77 

Construction.  Make  the  development  by  the  method  of  tri- 
angulation  as  explained  in  Article  107. 

142.  There  are  many  other  forms  of  warped  surfaces  than 
those  which  are  here  considered ;  but  as  they  are  seldom  encoun- 
tered in  practical  work  and  possess,  therefore,  little  interest  to 
the  engineer  or  architect,  it  has  been  thought  well  to  omit  them. 
The  principles  by  which  warped  surfaces  in  general  are  repre- 
sented and  by  which  problems  relating  to  such  surfaces  are  solved 
have  been  discussed  in  detail,  and  no  peculiar  difficulties  should 
be  encountered  in  solving  problems  relating  to  the  more  obscure 
surfaces  omitted  from  consideration  here. 


PROBLEMS  ON  WARPED  SURFACES 

257.  The  plane  T  is  o";  90;  45.  The  point  O  lies  in  plane  T  at  ij"; 
i\"\  o"  and  is  the  center  of  a  2"  circle  also  in  the  plane  T.  The  plane  S 
is  5";  90;  150.  The  point  Q  lies  in  the  H  trace  of  plane  S  in  the  profile 
plane  32";  90;  90  and  is  the  center  of  a  3''  circle  lying  in  plane  S.  With 
these  two  circles  as  directrices  and  the  V  plane  as  plane  director  draw  a  plan 
and  elevation  of  a  cylindroid. 

258.  The  plane  S  is  o";  90;  30.  The  point  0  is  in  the  H  trace  of  plane  S 
and  in  the  profile  plane  2^";  90;  90.  This  point  is  the  center  of  a  2\" 
circle  lying  in  plane  S.  The  line  MN  is  perpendicular  to  V  at  the  point 
A\"'i  o";  -"iJ"-  With  the  circle  and  line  as  directrices  and  the  V  plane  as 
plane  director  draw  a  plan,  elevation,  and  end  view  of  a  conoid. 

259.  The  point  A  is  o";  -3'';  o".  The  point  B  is  3";  o";  -3''.  The 
point  C  is  4";  o";  -^\".  The  point  D  is  5^";  -2,1";  o".  With  AB  and 
CD  as  directrices,  and  with  the  H  plane  as  plane  director  draw  the  plan, 
elevation,  and  end  view  of  the  hyperbolic  paraboloid. 

260.  A  cylinder  2\"  in  diameter  has  cut  on  it  a  single  square  thread  whose 
pitch  is  I".    Draw  a  plan  and  elevation. 

261.  A  4"  screw  has  3  standard  V-threads  per  inch.  Draw  a  plan  and 
elevation  of  the  screw. 

262.  A  bundle  conveyor  passes  from  the  top  floor  of  a  building  to  the 
wrapping  counter  in  the  basement.  The  distance  between  floors  is  10'  and 
in  passing  from  one  floor  to  the  floor  below  the  bundle  makes  3  turns.  The 
conveyor  is  in  the  form  of  a  right  helicoid  5'  in  diameter  and  mounted  on'a 
2'  shaft.     Draw  a  plan  and  elevation  of  the  conveyor. 

263.  Draw  the  plan  and  elevation  of  a  screw  conveyor  14''  in  diameter 
mounted  on  a  3"  shaft.  The  conveyor  is  capable  of  delivering  7500  cubic 
feet  per  hour  when  revolving  100  revolutions  per  minute. 


178  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

264.  Draw  the  plan  and  elevation  of  a  helical  spring  made  of  j''  square 
stock.    Inside  diameter  of  spring  2^";  length  6";  pitch  1". 

265.  Draw  the  plan  and  elevation  of  the  spring  in  Problem  264  made  up 
of  f "  round  stock. 

266.  A  conical  spring  is  6"  in  diameter  at  its  base  and  2"  in  diameter  at 
its  top.  It  is  made  up  of  4  turns  of  {"  square  stock.  Draw  the  plan  and 
elevation  of  the  spring. 

267.  Draw  a  plan  and  elevation  and  give  the  length  of  the  stock  required 
to  make  the  spring  in  Problem  266  out  of  f"  round  stock. 

268.  Draw  the  plan  and  elevation  of  a  convolute  conveyor  whose  diam- 
eter is  18",  whose  pitch  is  16'',  and  whose  shaft  is  6"  in  diameter. 

269.  Draw  a  plan  and  elevation  of  a  wood  screw  which  is  formed  by  two 
convolute  surfaces.  Outside  diameter  3'',  diameter  at  the  root  of  the 
thread  2",  pitch  2". 

270.  A  railway  which  runs  due  north  and  south  crosses  over  a  street 
which  runs  north  60  degrees  east  on  a  cow's  horn  arch.  The  openings  in  the 
arch  are  circles  40'  in  diameter  lying  in  parallel  planes  100'  apart.  Draw  a 
plan,  elevation  and  end  view  of  the  arch. 

271.  A  reducing  hood  whose  opening  is  a  circle  36"  in  diameter  and  whose 
outlet  is  a  circle  8''  in  diameter  is  fastened  to  a  wall  so  that  the  circular  open- 
ings are  tangent  to  the  wall.  These  tangent  points  are  in  a  line  which  is 
vertical  and  which  is  the  straight  line  directrix  of  the  surface.  The  length 
of  the  hood  along  the  wall  is  42'',  the  plane  of  the  opening  inclines  45  degrees 
to  the  wall,  and  the  plane  of  the  outlet  is  perpendicular  to  the  wall.  Draw  a 
plan  and  elevation  of  the  hood  and  make  a  pattern  for  ij;. 

272.  A  reducing  hood  has  two  circular  openings  whose  centers  are  in  a 
line  perpendicular  to  H,  and  36"  apart.  The  outlet  of  the  hood  is  a  circle 
18"  in  diameter  whose  plane  is  perpendicular  to  this  line;  and  the  inlet  is  a 
circle  whose  diameter  is  48"  and  whose  plane  incHnes  30  degrees  to  this 
line.    Draw  a  plan  and  elevation  of  this  hood  and  make  a  pattern  for  it. 

273.  Make  a  pattern  for  a  convolute  conveyor  flight  whose  diameter  is 
12",  pitch  8",  diameter  of  shaft  $". 

274.  Two  shafts,  one  vertical  and  one  inclined  45  degrees  in  a  vertical 
plane,  are  2''  in  diameter,  and  their  center  lines  are  7"  apart.  Draw  the  plan, 
elevation,  and  end  view  of  two  hyperboloids  of  revolution  which  will  operate 
on  these  shafts  in  the  ratio  of  4  to  3. 


CHAPTER   XVIII 


MODEL   MAKING 


143.  One  of  the  most  interesting  and  practical  means  of 
studying  the  development  of  surfaces  and  the  laying  out  of 
patterns  for  them  consists  in  the  construction  of  actual  scale 
models.  This  not  only  affords  practice  in  the  use  of  drawings 
and  gives  experience  in  constructing  the  object  represented  in 
the  drawing,  but  it  is  also  an  excellent  means  of  showing  how 
curves  of  intersection  appear  when  laid  on  a  pattern  as  well  as 


Fig.  143.  —  A  paper  model  of  a  blast  furnace  piping  problem  constructed  to  scale 
from  ordinary  detail  paper  and  mounted  on  a  wooden  base. 

when  they  are  an  actual  part  of  the  object  on  which  they  lie. 
While  such  model  making  must  of  course  be  done  to  scale  yet 
the  experience  acquired  serves  to  illustrate  many  of  the  practical 
considerations  which  must  be  kept  in  mind  when  laying  out 
sheet  metal  work. 

144.  Practical  Hints.  In  making  paper  models  such  as  the 
one  shown  in  Fig.  143  a  good  grade  of  detail  paper  will  give  satis- 
factory results.  After  the  developments  of  the  several  surfaces 
have  been  accurately  made  they  may  be  cut  out  —  leaving  an 
allowance  along  the  proper  edges  of  the  pattern  for  lap  —  and 
glued  together,  thus  forming  the  actual  surface. 

179 


i8o 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


Fig.  144.  —  illustration  showing  two  methods  of  making  joints  in  paper  model 
making.  The  method  at  the  right  gives  better  results  for  joints  along  straight 
lines  while  the  one  at  the  left  is  better  for  curved  joints. 


Fig.  144A.  —  Patterns  for  the  parts  of  the  model  shown  in  Fig.  147  before  being 
glued  together.    Note  the  method  for  making  the  joints. 


MODEL  MAKING  l8l 

Two  convenient  methods  for  making  the  joint  along  an  element 
of  the  surface  are  shown  in  Fig.  144.  The  joint  in  the  cylinder 
on  the  right  is  overlapped  and  on  the  one  side  of  the  pattern 
tongues  are  cut  to  fit  into  the  slits  on  the  other  side  of  the  pattern. 
This  method  has  been  found  to  give  good  results  and  is  some- 
what easier  than  the  one  shown  on  the  cylinder  at  the  left  in 
Fig.  144.  For  joints  of  two  intersecting  surfaces  it  will  be  found 
more  convenient  to  cut  tongues  on  both  patterns,  the  tongues  on 
one  side  being  arranged  to  fit  into  spaces  between  the  tongues 
on  the  other  as  shown  in  the  cylinder 
at  the  left  in  Fig.  144.  It  requires 
a  little  care  and  foresight  to  get  the 
tongues  and  spaces  to  knit  but  the 
resulting  joint  is  so  much  easier  to 
form  and  is  so  much  more  secure 
that  the  extra  trouble  is  worth  the 
time  it  takes. 

Both  Hbrary  paste  and  glue  are 

good   "solder"    for  the  joints. 

Library  paste  is  pleasanter  to  work 

with  and  takes  less  time  to  dry  but 

it  seems  to  hold  less  well  than  glue. 

A  dued  joint  has  been  found  to  last  ^^  '      T    '.       ,   ,         ,  , 

°  •'  Fig.  145. — A  view  of  the  model 

indefinitely  but  joints  fastened  with  shown  in  Fig.  146  cut  open  to 
paste  are  apt  to  break  open  in  time,      show  the  method  of  securing 

145.  The     tendency     of     surfaces       stiffness  and  true  form  by  insert- 

1        1.     1  1  ^S  cardboard  forms, 

such  as  cones  and  cylinders  when 

formed  from  detail  paper  patterns  is  to  be  flat  along  the 
joints.  This  is  due  to  the  extra  stiffness  at  these  points  caused 
by  overlapping  edges.  An  easy  method  of  overcoming  this  and 
securing  rigidity  in  the  model  is  to  insert  cardboard  reinforce- 
ments as  shown  in  Fig.  145.  These  cardboard  reinforcements 
may  be  made  in  the  form  of  right  sections  and  glued  in  place. 
The  model  will  thus  be  made  to  keep  its  proper  shape  and  will 
be  much  stiffer. 

146.  Naturally  a  model  such  as  the  one  shown  in  Fig.  146  will 
get  more  or  less  soiled  from  handling  during  its  construction  and 


l82 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


its  appearance  may  be  greatly  improved  if  upon  completion  it 
is  given  a  coat  of  bronze  or  silver  paint.  This  covers  the  glue 
marks  and  gives  the  model  a  fresh  and  finished  appearance  at  a 
trifling  cost. 


Fig.  146.  —  A  paper  model.  A  close 
examination  of  the  opening  in  the 
cowl  will  show  the  method  used 
to  make  the  joints. 


Fig.    147.  —  Another    paper 
model. 


Fig.  148.  —  A  collection  of  paper  models  made  by  the  methods  just  described. 
With  the  exception  of  two  or  three  all  of  these  models  were  made  by  students 
during  one  course. 


The  models  which  are  used  here  for  illustration,  both  the 
finished  models  and  the  ones  which  have  been  cut  open  to  show 
methods  of  construction,  were  made  by  students  in  the  regular 
required  course  in  descriptive  geometry  at  the  University  of 


MODEL   MAKING 


183 


Iowa.  This  fact  is  mentioned  to  impress  upon  students  who  use 
this  text  that  work  of  this  character  and  quahty  is  quite  within 
their  capacity. 

GENERAL  PROBLEMS 

275-288.  Draw  a  plan  and  elevation  of  the  object  shown  in  the  drawing 
and  lay  out  patterns  for  each  part.  After  the  patterns  have  been  laid  out 
to  the  same  scale  as  the  drawing,  they  are  to  be  fitted  together  and  glued  to 
form  a  model  of  the  object. 


Problem  275.    Hopper 


A 

B 

C 

D 

E 

F 

G 

a 

12" 

12" 

2" 

s" 

6" 

5" 

2" 

b 

36" 

18" 

6" 

0" 

18" 

0" 

6" 

^ 

36" 

18'' 

6" 

0" 

18" 

15" 

6" 

d 

36" 

18" 

6" 

6" 

18" 

0" 

6" 

e 

30'' 

12" 

A" 

4" 

12" 

9" 

X2" 

1 84  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


Problem  276.    Rectangular  Tapered  Connecting  Pipe 


A 

B 

C 

D 

E 

F 

G 

H 

L 

K 

a 

12" 

6" 

6" 

12" 

6" 

6" 

6" 

24" 

6" 

3" 

b 

12" 

3" 

6" 

9'' 

8" 

6" 

5" 

36" 

6" 

4^". 

c 

12" 

4" 

6" 

9" 

8" 

6" 

5" 

36" 

6" 

4" 

d 

12" 

6" 

6" 

12" 

6" 

6" 

6" 

24" 

6" 

6" 

e 

10" 

10" 

6" 

6" 

3" 

3" 

6'' 

20" 

6" 

0" 

MODEL  MAKING 


x8S 


Problem  277.    Twisted  Connecting  Pipe 


A 

B 

C 

D 

E 

F 

G 

a 

12" 

6" 

12" 

6" 

3" 

3" 

22" 

b 

18" 

12" 

20" 

9" 

6" 

6" 

30" 

c 

8" 

8" 

12" 

6'' 

4" 

4" 

24" 

d 

16" 

8" 

16" 

8" 

6" 

6" 

30'' 

e 

8" 

4" 

8" 

4"    • 

6" 

6" 

2A" 

i86 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


<      A 

i 

D 

c 

/ 

Problem  278.    Conical  Connection  for  Offset  Cylinders 


A 

B 

C 

D 

E 

X 

Notes 

a 

8'' 

16" 

8" 

8" 

14" 

60 

b 

12" 

18" 

12" 

10" 

12" 

45 

c 

8" 

16" 

8" 

8'' 

14" 

30 

d 

12" 

18" 

12" 

10'' 

12" 

30 

e 

8" 

14" 

10" 

8" 

.18'' 

60 

MODEL  MAKING 


187 


\ 

< F > 

^-^..^^ 

^^ 

i^f^ 

\ 

5 

^^.^ 

*                  U 

—1 

c 

3 

/ 

.    i 

N 

/ 
/ 
/ 

f 

3 

\ 

Problem  279.    Hopper  for  Pipe 


A 

B 

C 

D 

E 

F 

G 

X 

Pipe 
Section 

a 

12" 

6" 

6" 

10" 

3" 

s" 

10" 

90 

Rect. 

b 

15" 

IS" 

10" 

18" 

5" 

9" 

12" 

90 

Circle 

c 

6" 

6" 

6" 

10" 

3" 

5" 

10" 

90 

Circle 

d 

15" 

15" 

10" 

18" 

s" 

9" 

12" 

120 

Circle 

e 

9" 

9" 

6" 

10" 

3" 

s" 

10" 

135 

Circle 

Note:  In  all  of  the  above  problems  the  axis  of  the  hopper  remains  vertical. 


1 88  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


^ A > 


U — B  — 4 


Problem  280.    Conical  Hopper  on  Cylindrical  Pipe 


A 

B 

C 

D 

E 

X 

Y 

a 

12" 

12" 

14" 

12" 

12" 

90 

90 

b 

12" 

12" 

12" 

8" 

8" 

90 

90 

c 

18" 

12" 

20" 

12" 

18" 

75 

90 

d 

12" 

12" 

12" 

8" 

8" 

90 

75 

e 

12'' 

12" 

20" 

12" 

18" 

90 

90 

MODEL  MAKING 


189 


Problem  281.    Intersection  of  Cylindrical  Pipes 


A 

B 

C 

D 

E 

X 

Y 

a 

12" 

12" 

16" 

6" 

6" 

45 

90 

b 

12" 

18" 

20" 

6" 

10" 

60 

90 

c 

10" 

12" 

20" 

9" 

10" 

45 

90 

d 

10" 

12" 

20" 

9" 

10" 

60 

75 

e 

12" 

18" 

20" 

6" 

10" 

los 

90 

igo  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


Problem  282.    Reductng  Breeching 


MODEL  MAKING 


191 


mm. 


ff^yi/^^* 


jL 


<^^^ 


#^S5 00k 


^ 


iiH 


Problem  283.    Connecting  Rod  End 


A 

B 

C 

D 

E 

D' 

R 

a 

3" 

6" 

12" 

ir 

8" 

H" 

ir 

b 

4" 

8" 

12" 

4" 

10" 

5" 

l" 

c 

3" 

6" 

12" 

21" 

8" 

zY 

2" 

d 

4" 

8" 

10" 

2" 

10" 

3" 

A" 

e 

3" 

6" 

8" 

ir 

6" 

3" 

2" 

19^ 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


Problem  284.    Cowl  for  Ship's  Ventilator 


A 

B 

C 

D 

E 

X 

Number 
of  Sections 

a 

12" 

6" 

18'' 

8" 

14" 

105 

Five 

b 

18" 

9" 

24" 

9" 

18" 

105 

Five 

c 

18" 

9" 

24" 

9" 

18" 

105 

Four 

d 

12" 

6" 

18" 

8" 

14" 

90 

Two 

e 

14" 

8" 

20" 

9" 

12" 

105 

Five 

MODEL  MAKING 


193 


Problem  285.    Transition  Piece 


194 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


Problem  286.    Ventilator  Cowl  with  Y-Connection  and  Square  Hood 


A 

6" 
6" 
6" 

C 

D 

E 

F 

G 

H 

K 

L 
4" 

M 

N 

0 

8" 
8" 
8" 

X 

45 

Y 

Z 

a 

32" 

12" 

26" 

18" 

16" 

12" 

18" 

8" 

18" 
18" 
18" 

4" 
4" 

4S 
45 
60 

105 

b 

32" 

X2" 

26" 

18" 

16" 

12" 

18" 

8" 

4" 

60 

105 

c 

36" 

12" 

30" 

14" 

18" 

10" 

12" 

12" 

6" 

6" 

60 

105 

MODEL  MAKING 


195 


Problem  287.    Square  Hood  with  Y-Connection,  Elbow,  and 
Transition  Piece 


A 

B 

c 

D 

E 

F 

G 

8" 
8" 
8" 

H 

K 

L 

M 

8" 

N 

0 

P 

X 

Y 

Z 

a 

32- 

6" 
6" 

12" 

26" 

18" 

16" 

18" 

12" 

8" 

18" 

8'' 
8" 

8" 
6" 
8" 

45 

45 

60 

45 
45 

b 

32" 

12" 

26" 

18" 

16" 

20" 

10" 

8" 
8" 

8" 
8" 

18" 

60 

45 

c 

32" 

6" 

12" 

26" 

18" 

16" 

20" 

10" 

18" 

8" 

75 

60 

196 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


CHAPTER  XIX 


APPENDIX 

The  following  geometrical  constructions  will  be  found  useful 
in  drawing  many  of  the  problems  in  this  text.  While  it  is  under- 
stood that  many  of  them  are  well  known  problems  in  plane 
geometry  it  has  been  thought  well  to  include  them  here  for  the 
convenience  of  the  student. 

It  is  to  be  understood  that  no  attempt  has  been  made  to  prove 
any  of  the  constructions,  and  also  that  the  particular  construc- 
tion given  may  not  be  the  only 
method  of  arriving  at  the  same 
result. 

The  attempt  has  been  made  to 
draw  the  figure  in  such  a  way 
that  the  construction  will  be 
more  or  less  obvious,  and  there- 
fore the  explanation  has  been 
made  as  brief  as  possible.  The 
methods  here  given  have  been 
found  by  experience  to  be  the 
most  practical  constructions  for  ij{^ 
this  kind  of  work. 

Given  the  length  of  one  side  to  construct  a  regular  pentagon. 

Let  AB  be  the  given  side.  With  AB  as  a  radius  and  B  as  a 
center  draw  the  arc  NPAQC.  With  A  as  a  center  and  AB  as  the 
radius  draw  the  arc  MPBQE.  With  P  as  a  center  and  the  same 
radius  draw  the  arc  MAOBN.  Draw  the  lines  PQ,  MC,  NE, 
thus  locating  E  and  C.  With  E  and  C  as  centers  and  AB  as 
radius  draw  arcs  intersecting  at  D.  ABDCE,  then,  is  the  re- 
quired regular  pentagon. 

197 


198 


ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 


Given  the  length  of  one  side  to  construct  a  regular  hexagon. 

Let  AB  be  the  given  side.  With  A  and  B  as  centers  and  AB 
as  radii  draw  arcs  intersecting  at  0.  This  will  be  the  center  of 
the  hexagon.  With  O  as  a  center  and  a  radius  equal  to  OA  draw 
the  circumscribing  circle.  Lay  off  chords  on  this  circle  equal  in 
length  to  AB,  thus  locating  points  C,  D,  E,  and  F.    ABCDEF 

is  the  required  hexagon. 

E  P  D 

,      . — »^    ^_  IVI 7 ^?=*" 

E. 


Given  the  distance  "  across  the  flats  "  to  construct  a  regular 
hexagon. 

Let  MN  and  RS  be  two  lines  representing  the  sides  of  a  hexa- 
gon. Draw  PQ  perpendicular  to  MN  and  RS,  and  bisect  it  by 
the  perpendicular  TL.    The  point  0  is  the  center  of  the  inscribed 

circle  of  the  hexagon.  Draw 
OV  making  an  angle  of  60 
degrees  with  OQ.  Draw  also 
the  inscribed  circle  VQP  with 
O  as  a  center  and  OQ  as  a 
radius.  At  V  draw  AF  tan- 
gent to  OV;  AF  is  the  length 
of  one  side  of  the  required 
hexagon  which  may  now  be 
constructed  by  any  convenient 
method. 

Given  the  length  of  one  side  to  construct  a  regular  polygon 
of  any  given  number  of  sides. 

Let  AB  be  the  given  side  and  let  it  be  required  to  draw  a  poly- 
gon with  seven  sides.  With  A  as  a  center  and  a  radius  equal  to 
AB  draw  the  arc  BOGP.     Divide  this  arc  into  seven  equal  parts 


APPENDIX 


199 


and  through  the  second  point,  G,  draw  the  Kne  AG.  This  will 
be  the  side  of  the  polygon  adjacent  to  AB.  Bisect  AB  and  AG 
by  perpendiculars  meeting  at  0.  This  point  O  will  be  the  center 
of  the  circumscribing  circle.  Draw  this  circle  and  lay  off  on  it 
seven  chords  each  equal  to  AB,  thus  locating  the  points  C,  D,  E, 
and  F.  ABCDEFG, 
then,  is  the  required 
polygon  of  seven  sides. 

To  draw  a  tangent 
to  a  circle  from  a  given 
point  without  the  cir- 
cumference. 

Let  MANB  be  the 
given  circle  whose  cen- 
ter is  at  0,  and  let  P 

be  the  given  point  from  which  the  tangent  is  to  be  drawn.  Draw 
OP  and  upon  OP  as  a  diameter  describe  the  circle  OABP  cutting 
the  given  circle  at  the  points  A  and  B.  Draw  PA  and  PB ;  these 
will  be  the  required  tangents  from  P  and  A  and  B  will  be  the 
tangent  points. 


To  draw  a  line  tangent  to  two  given  circles. 

Let  MANB  be  one  circle  and  CD  the  other  circle  to  which  the 
line  is  to  be  drawn  tangent.  Draw  PO  connecting  the  centers. 
Lay  off  NI  equal  to  PD  and  describe  the  circle  HIJK.  This 
circle  is  equal  in  radius  to  the  difference  between  the  radii  of  the 
two  given  circles. 

Now  by  the  previous  problem  draw  the  tangents  PJ  and  PH 


200 


ESSENTIALS  OF  DESCRIPTIVE   GEOMETRY 


from  P  to  the  circle  HIJK.     Extend  OH  to  cut  the  given  circle 
at  A  and  extend  OJ  to  cut  at  B.    A  and  B  will  be  the  points  of 

tangency  and  AC  and  BD  drawn 
from  these  points  parallel  to 
HP  and  JB  will  be  the  two  lines 
which  may  be  drawn  tangent  to 
the  two  given  circles. 

To  draw  an  ellipse  by  tram- 
mels, given  the  axes. 

Let  MN  be  the  major  axis  and 
OP  the  minor  axis  of  the  ellipse. 
Cut  a  strip  of  paper  as  shown  and 
lay  off  AC  equal  to  half  of  MN 
and  lay  off  BC  equal  to  half  of  OP.  Now  keeping  the  point  A 
always  on  the  axis  OP  and  the  point  B  always  on  the  axis  MN 
move  the  trammel  about  and  mark  the  various  positions  of  C. 
A  sufficient  number  of  points  may  be  thus  marked  to  enable  the 
curve  to  be  accurately  drawn 
in  with  the  irregular  curve  as 
a  guide. 

To  locate  the  foci  of  the 
ellipse. 

With  P  as  a  center  and 
half  of  MN  as  a  radius  draw 
arcs  cutting  MN  at  F  and 
F',  the  required  foci. 

To  draw  a  tangent  to  an 
ellipse  at  a  point  on  the 
curve. 

Let  L  be  the  given  point. 
Draw  LF  and  LF',  the  focal 
radii.     Bisect    the    exterior 
angle  between  LF  and  LF'  and 
required  tangent. 

To  draw  an  ellipse,  given  the  axes.     Second  Method. 
Draw  the  circle  upon  MN,  the  major  axis;  and  draw  also  the 
circle  upon  PQ,  the  minor  axis.    Draw  oi,  02,  03,  etc.     From  a, 


APPENDIX 


20I 


b,  c,  etc.,  draw  horizontal  lines;  from  i,  2,  3,  etc.,  draw  vertical 
lines.  Locate  the  intersections  of  these  lines  as  at  i',  2',  3',  etc., 
and  find  a  sufficient  number  so  that  a  curve  may  be  drawn 
through  them  as  shown.     This  curve  will  be  the  required  elhpse. 

To  draw  a  tangent  to  an  ellipse  from  a  point  without  the  curve. 

Locate  F  and  F',  the  foci  of  the  given  eUipse.  Let  L  be  the 
given  point  at  which  the  tangent  is  to  be  drawn.  Draw  LF,  and 
with  LF  as  a  radius  and  L  as  a  center  describe  the  arc  FR.  With 
F'  as  a  center  and  MN  as  a  radius  describe  an  arc  cutting  the 
arc  FR  at  R,  and  draw  RF'.  The  point  where  this  line  RF'  cuts 
the  ellipse  or  K  is  the  point  of  tangency  and  LK,  then,  is  the 
required  tangent. 


To  draw  a  parabola  when  its  axis,  its  vertex,  and  a  point  on 
the  curve  are  given. 

Let  AB  be  the  axis,  A  the  vertex,  and  D  a  point  on  the  curve. 
Draw  DC  and  AC  and  divide  each  into  the  same  number  of  equal 
parts,  thus  locating  points  1,2,  etc.,  and  a,  b,  c,  etc.  Draw  Ai, 
A2,  A3,  A4,  A5,  A6,  A7,  and  find  where  each  of  these  lines  is 
intersected  by  lines  through  a,  b,  c,  d,  e,  f,  g  drawn  parallel  to 
AB.  The  intersections  thus  located  will  be  points  on  the  re- 
quired parabola,  and  when  a  sufficient  number  of  them  have 
been  found  the  curve  may  be  drawn  with  the  irregular  curve. 


202 


ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 


To  draw  a  tangent  to  a  parabola  at  a  point  on  the  curve. 

Let  L  be  the  given  point.  Draw  LP  perpendicular  to  AB  and 
make  NA  equal  to  PA.  Draw  NL;  this  will  be  the  required 
tangent  to  the  curve  at  L. 

To  draw  a  tangent  to  a  parabola  from  a  point  not  on  the 
curve. 

Let  CD  be  the  directrix  of  the  given  parabola  whose  focus  is 
F,  and  let  P  be  the  point  from  which  the  line  is  to  be  drawn  tan- 

C 


gent  to  the  curve.  With  P  as  a  center  and  a  radius  Pf  describe 
the  arc  cutting  the  directrix  at  K  and  L.  From  these  points 
draw  lines  LM  and  KN  perpendicular  to  CD  cutting  the  parab- 
ola at  M  and  N.  M  and  N,  then,  are  the  tangent  points,  and 
lines  drawn  from  P  to  these  points  will  be  tangent  to  the  curve. 

To  draw  an  hyperbola,  given  the  major  axis  and  one  point  on 
the  curve. 

Let  AB  be  the  given  axis  and  D  the  given  point.  Draw  the 
rectangle  BCDE,  and  divide  CD  and  ED  into  any  number  of 
equal  parts,  in  this  case  six.  Draw  A5,  A4,  A3,  A2,  Ai,  also 
B5,  B4,  B3,  B2,  Bi.  At  the  intersection  of  each  of  these  Hues 
from  A  with  the  corresponding  Unes  from  B  will  be  located  a 
point  on  the  curve,  and  by  finding  a  sufficient  number  of  points 
the  curve  may  be  drawn  as  shown. 


APPENDIX 


203 


To  draw  a  tangent  to  an  hyperbola  at  a  given  point  on  the  curve. 

Let  AB  be  the  major  axis  of  the  hyperbola  and  L  the  given 

point  at  which  the  tangent  is  to  be  drawn.     Draw  LM  perpen- 


dicular to  AB.  On  OM  as  a  diameter  describe  the  circle 
cutting  the  major  auxiliary  circle  at  K.  Draw  KN  perpen- 
dicular to  AB.     Draw  NL,  the  required  tangent. 

To  rectify  a  given  arc.     An  approximate  method. 

Let  AB  be  the  given  arc.  Divide  it  as  shown  into  a  number 
of  equal  parts.  Let  the  distances  into  which  the  arc  is  divided, 
as  Ai,  12,  23,  3B,  be  made  small  enough  so  that  the  chord  sub- 
tending this- small  arc  is 
practically  equal  to  the 
arc  itself.  With  the  bow 
dividers  lay  off  on  the 
given  straight  line  these 
distances.  The  distance 
AB  on  the  straight  line, 
then,  will  be  equal  to 
the  distance  AB  on  the 
arc. 

If  it  be  desired  to  lay  off  on  a  given  arc  any  given  length  the 
same  method  may  be  used.     The  whole  point  of  this  method 


204  ESSENTIALS  OF  DESCRIPTIVE  GEOMETRY 

is  in  keeping  the  distances  so  small  that  there  is  no  difference, 
practically,  between  the  arc  and  its  subtending  chord. 

Other  methods  of  laying  off  arcs  onto  straight  lines  may  be 
used,  but  the  above  is  by  far  the  most  convenient  for  the 
draftsman. 

To  measure  an  angle  by  its  natural  tangent. 

Let  the  given  angle  to  be  measured  be  ACB.     Its  natural 

AB 

tangent  is  — -.    Divide,  therefore,  AB  by  BC  and  look  up  the 
BC 

value  of  the  angle  corresponding  to  this  quotient  in  any  reliable 

^A  table  of  natural  tangents.  The 
result  will  be  the  value  of  the 
angle  in  degrees  and  minutes. 

To  lay  off  a  given  angle  by  its 
natural  tangent. 

Lay  off  B  any  convenient  dis- 
tance from  C,  say  3'',  and  erect 
a  perpendicular  BA.  To  find  the  length  of  BA  look  up  the  value 
of  the  natural  tangent  of  the  given  angle  in  any  reliable  table  of 
natural  tangents,  multiply  it  by  three,  and  lay  off  this  distance 
on  BA,  thus  locating  A.  Connect  A  with  C  and  BCA  then  will 
be  the  required  angle. 

Note.  Since  tables  of  natural  tangents  are  computed  on  a 
base  of  i  the  value  given  for  the  natural  tangent  must  always  be 
multiplied  by  the  length  of  the  base  when  it  is  taken  greater  than 
unity. 


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